Topological Spaces
Chapter 2 encoded nearness with a distance. Chapter 3 keeps the open-set structure and removes the numerical ruler. The topology axioms isolate exactly the operations needed to speak about local structure, closed sets, neighborhoods, and comparison of different notions of openness on the same underlying set.
Visual investigations · Before the formal course
Objectives
Move from metric examples to the abstract open-set structure of topology, and reason rigorously with the topology axioms.
- Verify whether a collection of subsets is a topology.
- Use open and closed sets as dual descriptions of the same structure.
- Compare topologies by inclusion and identify discrete, indiscrete, cofinite, and cocountable examples.
- Explain how a metric produces a topology while topology itself no longer requires a distance.
Prerequisites
Chapter 3 uses the set language of Chapter 1 and the metric-open-set intuition developed in Chapter 2.
- Arbitrary unions, finite intersections, complements, and De Morgan's laws.
- Proof by double inclusion and construction of counterexamples.
- Open balls and open sets in a metric space.
Notation and terminology
| \((X,\tau)\) | Topological space |
| \(U\in\tau\) | U is open |
| \(F\subseteq X\) | F is closed when \(X\setminus F\) is open |
| \(\tau_1\subseteq\tau_2\) | \(\tau_2\) is finer than \(\tau_1\) |
| \(\mathcal P(X)\) | Power set; the discrete topology |
| \(\tau_{\mathrm{ind}}\) | Indiscrete topology {\(\varnothing\),X} |
| \(\tau_{\mathrm{cof}}\) | Cofinite topology |
| \(\tau_d\) | Topology induced by a metric d |
Definitions
A topology on a set \(X\) is a collection \(\tau\subseteq\mathcal P(X)\) satisfying: \(\emptyset,X\in\tau\); arbitrary unions of members of \(\tau\) lie in \(\tau\); finite intersections of members of \(\tau\) lie in \(\tau\).
Let \(X=\{a,b,c\}\) and \(\tau=\{\varnothing,\{a\},\{a,b\},X\}\). First, \(\varnothing\) and \(X\) belong to \(\tau\), so (T1) holds. The two proper nonempty members are nested: \(\{a\}\subseteq\{a,b\}\). Therefore any union of members of \(\tau\) is again one of \(\varnothing,\{a\},\{a,b\},X\), and any finite intersection is also one of these four sets. Hence (T2) and (T3) hold, so \((X,\tau)\) is a topological space. This example illustrates the practical verification pattern: check the required sets, arbitrary unions, and finite intersections separately.
A subset \(F\subseteq X\) is closed when \(X\setminus F\) is open. A set may be open, closed, both, or neither.
Use \(X=\{a,b,c\}\) and \(\tau=\{\varnothing,\{a\},\{a,b\},X\}\). A set is closed exactly when its complement is in \(\tau\). Since \(X\setminus\{c\}=\{a,b\}\in\tau\), the singleton \(\{c\}\) is closed. Since \(X\setminus\{b,c\}=\{a\}\in\tau\), the set \(\{b,c\}\) is closed. Also \(\varnothing\) and \(X\) are both open and closed. By contrast, \(\{b\}\) is not closed because its complement \(\{a,c\}\) is not open. The example emphasizes that 'closed' does not mean 'not open'; openness and closedness are two independent properties linked by complementation.
For topologies \(\tau_1,\tau_2\) on the same set \(X\), \(\tau_2\) is finer than \(\tau_1\) when \(\tau_1\subseteq\tau_2\).
On \(X=\{a,b,c\}\), let \(\tau_0=\{\varnothing,X\}\), \(\tau_1=\{\varnothing,\{a\},X\}\), and \(\tau_2=\mathcal P(X)\). Then \(\tau_0\subsetneq\tau_1\subsetneq\tau_2\). Consequently \(\tau_1\) is finer than \(\tau_0\), while \(\tau_0\) is coarser than \(\tau_1\); similarly, the discrete topology \(\tau_2\) is finer than both. 'Finer' means that more subsets are declared open. It does not mean that the underlying set has changed: all three topologies live on the same set \(X\).
The indiscrete topology is \(\{\emptyset,X\}\), and the discrete topology is \(\mathcal P(X)\). The cofinite topology is \(\tau_{\mathrm{cof}}=\{\emptyset\}\cup\{U\subseteq X:X\setminus U\text{ is finite}\}\). The cocountable topology is \(\tau_{\mathrm{cc}}=\{\emptyset\}\cup\{U\subseteq X:X\setminus U\text{ is countable}\}\). Here “countable” includes finite sets.
Take \(X=\mathbb N\). The indiscrete topology contains only \(\varnothing\) and \(\mathbb N\), whereas the discrete topology contains every subset of \(\mathbb N\). The set \(U=\mathbb N\setminus\{2,5\}\) is cofinite-open because its complement is finite. Because \(\mathbb N\) itself is countable, every subset of \(\mathbb N\) has countable complement, so the cocountable topology on \(\mathbb N\) is actually discrete. On the uncountable set \(\mathbb R\), the behavior changes: \(\mathbb R\setminus\mathbb Q\) is cocountable-open because its complement \(\mathbb Q\) is countable, but it is not cofinite-open because \(\mathbb Q\) is infinite.
A set \(N\subseteq X\) is a neighborhood of \(x\in X\) if some open \(U\) satisfies \(x\in U\subseteq N\).
In \(\mathbb R\) with the usual topology, \([-1,1]\) is a neighborhood of \(0\): the open interval \((-1,1)\) satisfies \(0\in(-1,1)\subseteq[-1,1]\). The neighborhood itself is not open, because no usual-open interval around either endpoint is contained in \([-1,1]\). By contrast, \([0,1]\) is not a neighborhood of \(0\). Any usual-open set containing \(0\) contains some interval \((-\varepsilon,\varepsilon)\), and that interval contains negative points, so it cannot lie inside \([0,1]\). The witness open set in the definition is therefore essential.
For a metric space \((X,d)\), the metric topology \(\tau_d\) is the collection of all \(U\subseteq X\) such that, for every \(x\in U\), there exists \(r>0\) with \(B_d(x,r)\subseteq U\).
On \(\mathbb R\), use the metric \(d(x,y)=|x-y|\). Consider \(U=(0,2)\). If \(x\in U\), then both distances \(x-0\) and \(2-x\) are positive. Set \(r=\tfrac12\min\{x,2-x\}>0\). Whenever \(|y-x|<r\), the point \(y\) stays strictly between \(0\) and \(2\), so \(B_d(x,r)\subseteq U\). Hence \(U\in\tau_d\). In contrast, \([0,2]\) is not metric-open: at \(x=0\), every ball \((-r,r)\) contains negative points. This is the local-ball criterion in action.
Theorems and proofs
Let \((X,\tau)\) be a topological space. Its closed subsets contain \(\emptyset\) and \(X\), are closed under arbitrary intersections, and are closed under finite unions.
Proof. We translate each desired closed-set property into a statement about open complements. This works because, by Definition 3.2, a subset is closed exactly when its complement is open.
Step 1: the two distinguished sets. The complement of \(\varnothing\) is \(X\), and the complement of \(X\) is \(\varnothing\). Axiom (T1) says that both \(X\) and \(\varnothing\) are open. Therefore both \(\varnothing\) and \(X\) are closed.
Step 2: arbitrary intersections. Let \((F_i)_{i\in I}\) be any family of closed sets. For each \(i\), the complement \(X\setminus F_i\) is open. De Morgan's law gives \[X\setminus\bigcap_{i\in I}F_i=\bigcup_{i\in I}(X\setminus F_i).\] The right-hand side is an arbitrary union of open sets, so it is open by (T2). Hence the complement of \(\bigcap_{i\in I}F_i\) is open, which means \(\bigcap_{i\in I}F_i\) is closed. If \(I=\varnothing\), the intersection is \(X\), already covered by Step 1.
Step 3: finite unions. Let \(F_1,\ldots,F_n\) be closed. Again each \(X\setminus F_j\) is open, and De Morgan's law gives \[X\setminus\bigcup_{j=1}^{n}F_j=\bigcap_{j=1}^{n}(X\setminus F_j).\] The right-hand side is a finite intersection of open sets, so it is open by (T3). Therefore \(\bigcup_{j=1}^{n}F_j\) is closed. The empty finite union is \(\varnothing\), which is closed by Step 1.
We have proved exactly the three closed-set axioms: \(\varnothing,X\) are closed, arbitrary intersections of closed sets are closed, and finite unions of closed sets are closed. \(\square\)
Let \(\{\tau_\alpha\}_{\alpha\in A}\) be a nonempty family of topologies on the same set \(X\). Then \[\tau=\bigcap_{\alpha\in A}\tau_\alpha\] is a topology on \(X\). By contrast, the union of two topologies need not be a topology.
Proof. Put \(\tau=\bigcap_{\alpha\in A}\tau_\alpha\). By definition, a subset \(U\subseteq X\) belongs to \(\tau\) exactly when it belongs to every topology \(\tau_\alpha\). We verify the three topology axioms for this common part.
Step 1: (T1). Every \(\tau_\alpha\) contains \(\varnothing\) and \(X\). Hence these two sets belong simultaneously to all \(\tau_\alpha\), and therefore \(\varnothing,X\in\tau\).
Step 2: (T2). Let \((U_i)_{i\in I}\) be any family of sets in \(\tau\). Fix \(\alpha\in A\). Since every \(U_i\in\tau\), every \(U_i\in\tau_\alpha\). Because \(\tau_\alpha\) is a topology, \(\bigcup_{i\in I}U_i\in\tau_\alpha\). The choice of \(\alpha\) was arbitrary, so this union belongs to every \(\tau_\alpha\), hence to their intersection \(\tau\).
Step 3: (T3). If \(U_1,\ldots,U_n\in\tau\), then for every \(\alpha\) all of these sets lie in \(\tau_\alpha\). Finite-intersection closure in \(\tau_\alpha\) gives \(\bigcap_{j=1}^{n}U_j\in\tau_\alpha\). Since this is true for every \(\alpha\), the intersection lies in \(\tau\). Thus \(\tau\) is a topology.
Why unions of topologies can fail. On \(X=\{a,b,c\}\), the collections \(\tau_1=\{\varnothing,\{a\},X\}\) and \(\tau_2=\{\varnothing,\{b\},X\}\) are topologies. Their set-theoretic union contains \(\{a\}\) and \(\{b\}\), but it does not contain their union \(\{a,b\}\). Therefore \(\tau_1\cup\tau_2\) violates (T2) and is not a topology. \(\square\)
For every set \(X\), the cofinite and cocountable collections are topologies. If \(X\) is finite, the cofinite topology is discrete. If \(X\) is countable, the cocountable topology is discrete.
Proof. We verify the topology axioms first for the cofinite collection and then explicitly for the cocountable collection. The empty set is included by definition in both collections.
Cofinite topology, (T1). We have \(\varnothing\in\tau_{\mathrm{cof}}\). Also \(X\in\tau_{\mathrm{cof}}\) because \(X\setminus X=\varnothing\) is finite.
Cofinite topology, arbitrary unions. Let \((U_i)_{i\in I}\subseteq\tau_{\mathrm{cof}}\). If every \(U_i\) is empty, their union is \(\varnothing\). Otherwise choose \(j\) with \(U_j\ne\varnothing\). Then \[X\setminus\bigcup_{i\in I}U_i\subseteq X\setminus U_j.\] The set on the right is finite, and every subset of a finite set is finite. Hence the complement of the union is finite, so the union is cofinite-open.
Cofinite topology, finite intersections. Let \(U_1,\ldots,U_n\) be cofinite-open. If one of them is empty, their intersection is empty. Otherwise each complement \(X\setminus U_k\) is finite, and \[X\setminus\bigcap_{k=1}^{n}U_k=\bigcup_{k=1}^{n}(X\setminus U_k).\] A finite union of finite sets is finite, so the intersection is cofinite-open.
Cocountable topology. The same structure must be checked with 'countable' in place of 'finite'. For arbitrary unions, if the union is nonempty choose a nonempty member \(U_j\); the complement of the whole union is contained in the countable set \(X\setminus U_j\), hence is countable. For a finite intersection of nonempty cocountable-open sets, De Morgan gives a finite union of countable complements, which is countable. Thus \(\tau_{\mathrm{cc}}\) is also a topology.
Special cases. If \(X\) is finite, every subset \(U\subseteq X\) has finite complement, so \(\tau_{\mathrm{cof}}=\mathcal P(X)\). If \(X\) is countable, every subset has countable complement, so \(\tau_{\mathrm{cc}}=\mathcal P(X)\). In the respective cases the topology is discrete. \(\square\)
Let \((X,d)\) be a metric space. The collection \(\tau_d\) of metric-open subsets of \(X\) is a topology on \(X\).
Proof. Recall the definition: a subset \(U\subseteq X\) is metric-open if every point \(x\in U\) has some radius \(r>0\) for which \(B_d(x,r)\subseteq U\). We verify (T1), (T2), and (T3) directly from this local condition.
Step 1: (T1). The empty set is metric-open because there is no point \(x\in\varnothing\) for which the ball condition could fail. The whole set \(X\) is metric-open because, for any \(x\in X\), any radius \(r>0\) satisfies \(B_d(x,r)\subseteq X\).
Step 2: arbitrary unions. Let \((U_\alpha)_{\alpha\in A}\) be metric-open and put \(U=\bigcup_{\alpha\in A}U_\alpha\). Take an arbitrary point \(x\in U\). By the meaning of union, some index \(\beta\) satisfies \(x\in U_\beta\). Since \(U_\beta\) is metric-open, there exists \(r>0\) with \(B_d(x,r)\subseteq U_\beta\). Because \(U_\beta\subseteq U\), we have \(B_d(x,r)\subseteq U\). This works for every \(x\in U\), so \(U\) is metric-open.
Step 3: two-set intersections. Let \(U,V\) be metric-open and take \(x\in U\cap V\). Choose \(r>0\) with \(B_d(x,r)\subseteq U\), and \(s>0\) with \(B_d(x,s)\subseteq V\). Put \(\rho=\min\{r,s\}>0\). Then \(B_d(x,\rho)\subseteq B_d(x,r)\cap B_d(x,s)\subseteq U\cap V\). Thus \(U\cap V\) is metric-open.
Step 4: all finite intersections. Repeating the two-set argument, or taking the minimum of finitely many positive radii, proves closure under every nonempty finite intersection. The empty finite intersection is \(X\), already open by Step 1. Therefore \(\tau_d\) satisfies all topology axioms and is a topology on \(X\). \(\square\)
Different metrics can induce the same topology. From this chapter onward, open sets are primary unless a metric is explicitly supplied.
Worked examples
For \(X=\{a,b\}\), exactly four topologies exist: \(\{\emptyset,X\}\), \(\{\emptyset,\{a\},X\}\), \(\{\emptyset,\{b\},X\}\), and \(\mathcal P(X)\).
Every cofinite subset of \(\mathbb R\) is standard-open, so \(\tau_{\mathrm{cof}}\subsetneq\tau_{\mathrm{std}}\).
In \(\mathbb R\), \([-1,1]\) is a neighborhood of \(0\) because it contains \((-1,1)\), but it is not open.
Exercises
Work through each exercise in the LaTeX workspace below. Use the progressive hints only after a genuine attempt. When you reveal a correction, read the goal and reasoning plan first, then compare each step of the complete solution with your own argument.
Work through each exercise in the LaTeX workspace. Use the three progressive hints only after a genuine attempt.
Expected evidence: A topology-axiom verification that explicitly checks (T1), arbitrary unions, and finite intersections for the four listed subsets.
1. Goal.
On \(X=\{1,2,3\}\), determine whether \(\tau=\{\varnothing,\{1\},\{1,2\},X\}\) is a topology.
2. Reasoning plan.
Checkpoint 1. Start with (T1): verify that both \(\varnothing\) and \(X\) appear in the collection.
Checkpoint 2. Notice that the collection is a chain \(\varnothing\subseteq\{1\}\subseteq\{1,2\}\subseteq X\). For a nested family, every union is its largest participating member.
Checkpoint 3. Use the same nesting observation for finite intersections: every nonempty finite intersection is the smallest participating member; the empty finite intersection is \(X\).
3. Learning check before the full solution.
A common mistake is to check only pairwise unions. The axiom asks for arbitrary unions, so explain why every subfamily has a union still in the collection.
4. Complete step-by-step solution.
Proof. Let \(\tau=\{\varnothing,\{1\},\{1,2\},X\}\), where \(X=\{1,2,3\}\). First, \(\varnothing,X\in\tau\), so (T1) holds. For (T2), take any subfamily \(\mathcal U\subseteq\tau\). Because the four members are linearly ordered by inclusion, if \(\mathcal U\) is nonempty its union is simply the largest set occurring in \(\mathcal U\); therefore it is one of the four members of \(\tau\). If \(\mathcal U\) is empty, its union is \(\varnothing\in\tau\). For (T3), any nonempty finite intersection of members of this chain is the smallest participating member, hence lies in \(\tau\); the empty finite intersection is \(X\in\tau\). All three axioms hold. Therefore \(\tau\) is a topology on \(X\). \(\square\)
Expected evidence: An exhaustive classification: explain why only the two singletons are optional, list the four possibilities, and verify that each is a topology.
1. Goal.
List all topologies on \(X=\{a,b\}\) and prove that your list is complete.
2. Reasoning plan.
Checkpoint 1. Every topology on \(X=\{a,b\}\) must contain \(\varnothing\) and \(X\).
Checkpoint 2. The power set has only two additional subsets, \(\{a\}\) and \(\{b\}\), so classify according to which of these are included.
Checkpoint 3. After listing the four cases, verify each one and explain why no fifth case can exist.
3. Learning check before the full solution.
Completeness is part of the exercise. Listing four valid examples is not enough unless you show that every topology must be one of them.
4. Complete step-by-step solution.
Solution. The power set is \(\mathcal P(X)=\{\varnothing,\{a\},\{b\},X\}\). A topology must contain \(\varnothing\) and \(X\), so the only choices concern the two singletons. If neither singleton is included, we obtain \(\tau_0=\{\varnothing,X\}\). If only \(\{a\}\) is included, we obtain \(\tau_a=\{\varnothing,\{a\},X\}\). If only \(\{b\}\) is included, we obtain \(\tau_b=\{\varnothing,\{b\},X\}\). If both are included, their union is \(X\), already present, and we obtain \(\mathcal P(X)\). Each of these collections is closed under arbitrary unions and finite intersections; in the three smaller cases the members are nested, and the power set is automatically closed under unions and intersections. These four choices exhaust all possibilities for the two optional subsets, so there are exactly four topologies on \(X\).
Expected evidence: Separate proofs that the indiscrete and discrete collections satisfy (T1)-(T3), followed by both inclusions for an arbitrary topology.
1. Goal.
Prove that the indiscrete and discrete collections are topologies on every set \(X\), and that every topology \(\tau\) on \(X\) satisfies \[\{\varnothing,X\}\subseteq\tau\subseteq\mathcal P(X).\]
2. Reasoning plan.
Checkpoint 1. For the indiscrete collection, the only possible unions or intersections of members are again \(\varnothing\) or \(X\).
Checkpoint 2. For the discrete collection \(\mathcal P(X)\), any union or intersection of subsets of \(X\) is still a subset of \(X\).
Checkpoint 3. For an arbitrary topology \(\tau\), use (T1) for the left inclusion and \(\tau\subseteq\mathcal P(X)\) from the meaning of 'collection of subsets' for the right inclusion.
3. Learning check before the full solution.
The extremal inclusions say more than validity: they identify the indiscrete topology as the coarsest and the discrete topology as the finest topology on the fixed set \(X\).
4. Complete step-by-step solution.
Proof. Let \(\tau_{\mathrm{ind}}=\{\varnothing,X\}\). It contains the two required sets. An arbitrary union of members is either \(\varnothing\) or \(X\), and a finite intersection is also either \(\varnothing\) or \(X\); hence it is a topology. Now let \(\tau_{\mathrm{disc}}=\mathcal P(X)\). It contains \(\varnothing\) and \(X\), and unions and finite intersections of subsets of \(X\) are still subsets of \(X\), so they remain in \(\mathcal P(X)\). Thus it too is a topology. Finally, if \(\tau\) is any topology on \(X\), axiom (T1) gives \(\{\varnothing,X\}\subseteq\tau\). Since every member of \(\tau\) is, by definition, a subset of \(X\), we also have \(\tau\subseteq\mathcal P(X)\). Therefore \(\tau_{\mathrm{ind}}\subseteq\tau\subseteq\tau_{\mathrm{disc}}\).
Expected evidence: A complement-based derivation of all three closed-set axioms using De Morgan's laws, including the arbitrary-intersection and finite-union cases.
1. Goal.
Derive the three closed-set axioms from the open-set axioms using complements.
2. Reasoning plan.
Checkpoint 1. Translate 'closed' into 'open complement'.
Checkpoint 2. For arbitrary intersections of closed sets, apply De Morgan to turn the complement into an arbitrary union of open sets.
Checkpoint 3. For finite unions of closed sets, apply De Morgan to turn the complement into a finite intersection of open sets.
3. Learning check before the full solution.
Keep the direction of De Morgan's laws straight: complementation reverses unions and intersections, and the topology axioms have different allowed cardinalities for those operations.
4. Complete step-by-step solution.
Proof. First, \(\varnothing\) is closed because \(X\setminus\varnothing=X\) is open, and \(X\) is closed because \(X\setminus X=\varnothing\) is open. Next let \((F_i)_{i\in I}\) be any family of closed sets. Then each \(X\setminus F_i\) is open and \[X\setminus\bigcap_{i\in I}F_i=\bigcup_{i\in I}(X\setminus F_i).\] The right-hand side is open by arbitrary-union closure, so \(\bigcap_iF_i\) is closed. Finally, for closed \(F_1,\ldots,F_n\), \[X\setminus\bigcup_{j=1}^{n}F_j=\bigcap_{j=1}^{n}(X\setminus F_j).\] The right-hand side is open by finite-intersection closure, so the finite union is closed. These are exactly the three closed-set axioms. \(\square\)
Expected evidence: A direct verification of (T1)-(T3) for the common sets in the intersection, with quantifiers made explicit.
1. Goal.
Let \(\{\tau_\alpha\}_{\alpha\in A}\) be a nonempty family of topologies on X. Prove that \(\bigcap_{\alpha\in A}\tau_\alpha\) is a topology.
2. Reasoning plan.
Checkpoint 1. Membership in \(\bigcap_\alpha\tau_\alpha\) means membership in every \(\tau_\alpha\).
Checkpoint 2. Check arbitrary unions after fixing an arbitrary \(\alpha\); the union is open in that topology because all its members are.
Checkpoint 3. Repeat for finite intersections and then release the arbitrary index \(\alpha\).
3. Learning check before the full solution.
The proof works because the same union or intersection is formed in every topology. Do not confuse intersection of topologies, whose elements are subsets of \(X\), with intersection of open subsets of \(X\).
4. Complete step-by-step solution.
Proof. Let \(\tau=\bigcap_{\alpha\in A}\tau_\alpha\). Because every \(\tau_\alpha\) is a topology, \(\varnothing\) and \(X\) belong to every \(\tau_\alpha\); hence they belong to \(\tau\). Now let \((U_i)_{i\in I}\subseteq\tau\). For each \(\alpha\), every \(U_i\in\tau_\alpha\). Therefore \(\bigcup_{i\in I}U_i\in\tau_\alpha\) by (T2). Since this holds for every \(\alpha\), the union belongs to \(\tau\). Likewise, if \(U_1,\ldots,U_n\in\tau\), then for each \(\alpha\) the finite intersection \(\bigcap_{j=1}^{n}U_j\in\tau_\alpha\), so it lies in \(\tau\). Thus \(\tau\) satisfies (T1)-(T3) and is a topology. \(\square\)
Expected evidence: Two explicit topologies and one specific union of open sets that is missing from their set-theoretic union.
1. Goal.
Give two topologies on the same set whose union is not a topology.
2. Reasoning plan.
Checkpoint 1. Choose a three-point set so two distinct singletons can be open without forcing their union to be listed in either topology.
Checkpoint 2. Verify separately that each proposed collection is a topology.
Checkpoint 3. In the set-theoretic union, identify two members whose union is missing; that single failure disproves the topology axiom.
3. Learning check before the full solution.
A counterexample to 'the union of topologies is a topology' requires the original collections to be genuine topologies. Verify that before showing the failure of their union.
4. Complete step-by-step solution.
Counterexample. Let \(X=\{a,b,c\}\), \(\tau_1=\{\varnothing,\{a\},X\}\), and \(\tau_2=\{\varnothing,\{b\},X\}\). In each collection the members are nested, so arbitrary unions and finite intersections remain in the collection; both are topologies. Their set-theoretic union is \[\tau_1\cup\tau_2=\{\varnothing,\{a\},\{b\},X\}.\] It contains the two open sets \(\{a\}\) and \(\{b\}\). However, topology axiom (T2) would require their union \(\{a,b\}\) to belong to \(\tau_1\cup\tau_2\), and it does not. Therefore \(\tau_1\cup\tau_2\) is not a topology. This pinpoints the exact failure: arbitrary-union closure. \(\square\)
Expected evidence: A full topology-axiom proof for the cofinite collection, treating the empty-union case and using complements for finite intersections.
1. Goal.
Prove that the cofinite collection \(\tau_{\mathrm{cof}}=\{\varnothing\}\cup\{U\subseteq X:X\setminus U\text{ is finite}\}\) is a topology on X.
2. Reasoning plan.
Checkpoint 1. Check \(\varnothing\) directly and check \(X\) through its finite complement \(\varnothing\).
Checkpoint 2. For an arbitrary union, separate the all-empty case; otherwise its complement is contained in the finite complement of one nonempty member.
Checkpoint 3. For a finite intersection, use De Morgan: the complement is a finite union of finite complements.
3. Learning check before the full solution.
The arbitrary-union argument does not say an arbitrary union of finite sets is finite. It controls the complement of the union by inclusion in one finite complement.
4. Complete step-by-step solution.
Proof. By definition \(\varnothing\in\tau_{\mathrm{cof}}\), and \(X\in\tau_{\mathrm{cof}}\) because \(X\setminus X=\varnothing\) is finite. Let \((U_i)_{i\in I}\) be cofinite-open. If every \(U_i=\varnothing\), then the union is \(\varnothing\), hence open. Otherwise choose \(j\) with \(U_j\ne\varnothing\). Then \[X\setminus\bigcup_{i\in I}U_i\subseteq X\setminus U_j.\] The right-hand side is finite, so the left-hand side is finite; therefore the union is cofinite-open. Now let \(U_1,\ldots,U_n\) be cofinite-open. If some \(U_k=\varnothing\), the intersection is \(\varnothing\). Otherwise each \(X\setminus U_k\) is finite and \[X\setminus\bigcap_{k=1}^{n}U_k=\bigcup_{k=1}^{n}(X\setminus U_k),\] a finite union of finite sets, hence finite. Thus all topology axioms hold. \(\square\)
Expected evidence: A pointwise argument that every subset of a finite set has finite complement, followed by the identification with the power set.
1. Goal.
If X is finite, prove that its cofinite topology is the discrete topology.
2. Reasoning plan.
Checkpoint 1. Take an arbitrary subset \(U\subseteq X\).
Checkpoint 2. Because \(X\) is finite, every subset of \(X\), including \(X\setminus U\), is finite.
Checkpoint 3. Therefore every \(U\subseteq X\) is cofinite-open; compare the resulting collection with \(\mathcal P(X)\).
3. Learning check before the full solution.
To prove equality with the discrete topology, show that every subset belongs to the cofinite topology, not merely that many examples do.
4. Complete step-by-step solution.
Proof. Suppose \(X\) is finite. Let \(U\subseteq X\) be arbitrary. Its complement \(X\setminus U\) is a subset of the finite set \(X\), so \(X\setminus U\) is finite. If \(U=\varnothing\), it belongs to \(\tau_{\mathrm{cof}}\) by definition; if \(U\ne\varnothing\), the finiteness of its complement places it in \(\tau_{\mathrm{cof}}\). Thus every subset \(U\subseteq X\) belongs to \(\tau_{\mathrm{cof}}\), so \(\mathcal P(X)\subseteq\tau_{\mathrm{cof}}\). The reverse inclusion is automatic because a topology on \(X\) is a collection of subsets of \(X\). Hence \(\tau_{\mathrm{cof}}=\mathcal P(X)\), which is exactly the discrete topology. \(\square\)
Expected evidence: A full topology-axiom proof for the cocountable collection, with the facts about subsets and finite unions of countable sets stated explicitly.
1. Goal.
Prove that the cocountable collection \(\tau_{\mathrm{cc}}=\{\varnothing\}\cup\{U\subseteq X:X\setminus U\text{ is countable}\}\) is a topology.
2. Reasoning plan.
Checkpoint 1. Verify \(\varnothing\) and \(X\) first.
Checkpoint 2. For an arbitrary nonempty union, its complement is contained in the countable complement of one nonempty member.
Checkpoint 3. For a finite intersection, De Morgan produces a finite union of countable sets, which is countable.
3. Learning check before the full solution.
Two countability facts are being used: every subset of a countable set is countable, and every finite union of countable sets is countable.
4. Complete step-by-step solution.
Proof. By definition \(\varnothing\in\tau_{\mathrm{cc}}\), and \(X\in\tau_{\mathrm{cc}}\) because \(X\setminus X=\varnothing\) is countable. Let \((U_i)_{i\in I}\subseteq\tau_{\mathrm{cc}}\). If every \(U_i\) is empty, their union is empty. Otherwise choose \(j\) with \(U_j\ne\varnothing\). Then \[X\setminus\bigcup_iU_i\subseteq X\setminus U_j.\] The right side is countable, so its subset on the left is countable; hence the union is cocountable-open. For a finite family \(U_1,\ldots,U_n\), an empty member makes the intersection empty. Otherwise \[X\setminus\bigcap_{k=1}^{n}U_k=\bigcup_{k=1}^{n}(X\setminus U_k).\] Each complement is countable, and a finite union of countable sets is countable. Therefore the intersection is cocountable-open. All three topology axioms hold. \(\square\)
Expected evidence: Proofs of both strict inclusions and two counterexamples establishing that the cocountable and usual topologies are incomparable.
1. Goal.
On \(\mathbb R\), compare the cofinite, cocountable, and usual topologies. Prove \(\tau_{\mathrm{cof}}\subsetneq\tau_{\mathrm{cc}}\) and \(\tau_{\mathrm{cof}}\subsetneq\tau_{\mathrm{std}}\), and show \(\tau_{\mathrm{cc}}\) and \(\tau_{\mathrm{std}}\) are incomparable.
2. Reasoning plan.
Checkpoint 1. Finite sets are countable, so cofinite-open implies cocountable-open.
Checkpoint 2. Finite subsets of \(\mathbb R\) are usual-closed, so their complements are usual-open.
Checkpoint 3. Use \(\mathbb R\setminus\mathbb Q\) and \((0,1)\) both to prove strictness and to show the latter two topologies are incomparable.
3. Learning check before the full solution.
For incomparability, you need one set open in the cocountable topology but not in the usual topology, and another set open in the usual topology but not cocountable-open.
4. Complete step-by-step solution.
Solution. If \(U\in\tau_{\mathrm{cof}}\) and \(U\ne\varnothing\), then \(\mathbb R\setminus U\) is finite, hence countable; therefore \(U\in\tau_{\mathrm{cc}}\). Thus \(\tau_{\mathrm{cof}}\subseteq\tau_{\mathrm{cc}}\). The inclusion is strict because \(\mathbb R\setminus\mathbb Q\) has countable complement \(\mathbb Q\), so it is cocountable-open, but the complement is infinite, so it is not cofinite-open. Also, a finite subset of \(\mathbb R\) is usual-closed, hence every cofinite set is usual-open. Thus \(\tau_{\mathrm{cof}}\subseteq\tau_{\mathrm{std}}\), strictly because \((0,1)\) is usual-open but has infinite complement. Finally, \(\mathbb R\setminus\mathbb Q\) is not usual-open since every interval about an irrational contains rationals, while \((0,1)\) is not cocountable-open because its complement is uncountable. Hence \(\tau_{\mathrm{cc}}\) and \(\tau_{\mathrm{std}}\) are incomparable.
Expected evidence: Verification that the proposed collection is a topology, then a complete closed-set list obtained by taking complements of every open set.
1. Goal.
On \(X=\{a,b,c\}\), verify \(\tau=\{\varnothing,\{a\},\{a,b\},X\}\) and list all closed sets.
2. Reasoning plan.
Checkpoint 1. Verify the topology exactly as in Exercise 3.1: the members are nested.
Checkpoint 2. Closed sets are complements of open sets; compute the complement of each of the four open sets.
Checkpoint 3. Check that the list has four closed sets and pair each with its open complement.
3. Learning check before the full solution.
Do not infer the closed sets by appearance. In an arbitrary topology, closedness is defined by the complement being open.
4. Complete step-by-step solution.
Solution. The collection is \(\tau=\{\varnothing,\{a\},\{a,b\},X\}\). Since \(\varnothing\subseteq\{a\}\subseteq\{a,b\}\subseteq X\), arbitrary unions and finite intersections of members stay in the chain, and \(\varnothing,X\) are present. Hence \(\tau\) is a topology. Now compute complements in \(X=\{a,b,c\}\): \(X\setminus\varnothing=X\), \(X\setminus\{a\}=\{b,c\}\), \(X\setminus\{a,b\}=\{c\}\), and \(X\setminus X=\varnothing\). Therefore the complete family of closed sets is \[\{X,\{b,c\},\{c\},\varnothing\}.\] There are no other closed sets because a set is closed exactly when its complement is one of the four open sets just listed. \(\square\)
Expected evidence: The strict inclusion between the two collections translated correctly into the finer/coarser terminology.
1. Goal.
Let \(\tau_1=\{\varnothing,X\}\) and \(\tau_2=\{\varnothing,\{a\},X\}\) on \(X=\{a,b\}\). Determine which is finer and which is coarser.
2. Reasoning plan.
Checkpoint 1. Compare the collections as sets of subsets.
Checkpoint 2. Every member of \(\tau_1\) appears in \(\tau_2\), and \(\{a\}\) shows the inclusion is strict.
Checkpoint 3. Translate \(\tau_1\subsetneq\tau_2\) using the definition of finer and coarser.
3. Learning check before the full solution.
Fineness is ordered in the same direction as inclusion: the topology with more open sets is finer.
4. Complete step-by-step solution.
Solution. We have \(\tau_1=\{\varnothing,X\}\) and \(\tau_2=\{\varnothing,\{a\},X\}\). Both members of \(\tau_1\) occur in \(\tau_2\), so \(\tau_1\subseteq\tau_2\). The subset \(\{a\}\) belongs to \(\tau_2\) but not to \(\tau_1\), so the inclusion is strict: \(\tau_1\subsetneq\tau_2\). By Definition 3.3, when \(\tau_1\subseteq\tau_2\), the topology \(\tau_2\) is finer than \(\tau_1\), and \(\tau_1\) is coarser than \(\tau_2\). Because the inclusion is strict here, we may say \(\tau_2\) is strictly finer and \(\tau_1\) strictly coarser. The underlying set \(X\) is unchanged; only the family of declared open subsets has increased.
Expected evidence: A proof that every cofinite real subset is usual-open, followed by the set-theoretic consequence for the intersection of topologies.
1. Goal.
On \(\mathbb R\), prove \(\tau_{\mathrm{cof}}\cap\tau_{\mathrm{std}}=\tau_{\mathrm{cof}}\).
2. Reasoning plan.
Checkpoint 1. Show first that \(\tau_{\mathrm{cof}}\subseteq\tau_{\mathrm{std}}\).
Checkpoint 2. Use the usual fact that finite subsets of \(\mathbb R\) are closed, so their complements are open.
Checkpoint 3. Once one family is contained in the other, their intersection is the smaller family.
3. Learning check before the full solution.
The identity \(A\cap B=A\) follows from \(A\subseteq B\). The mathematical work is therefore proving the inclusion of the two topologies.
4. Complete step-by-step solution.
Proof. Let \(U\in\tau_{\mathrm{cof}}\). If \(U=\varnothing\), then \(U\) is usual-open. Otherwise \(F=\mathbb R\setminus U\) is finite. Every singleton \(\{x\}\) is closed in the usual topology on \(\mathbb R\), and a finite union of closed sets is closed, so the finite set \(F\) is closed. Hence its complement \(U=\mathbb R\setminus F\) is usual-open. Thus every cofinite-open set is usual-open, giving \(\tau_{\mathrm{cof}}\subseteq\tau_{\mathrm{std}}\). For any two sets \(A\subseteq B\), one has \(A\cap B=A\). Applying this to the two topologies yields \[\tau_{\mathrm{cof}}\cap\tau_{\mathrm{std}}=\tau_{\mathrm{cof}}.\] \(\square\)
Expected evidence: A proof of the exact infinite intersection, followed by an explicit reason that the singleton \(\{0\}\) is not usual-open.
1. Goal.
In the usual topology on \(\mathbb R\), show that arbitrary intersections of open sets need not be open by using \(U_n=(-1/n,1/n)\).
2. Reasoning plan.
Checkpoint 1. First show \(0\) belongs to every interval \((-1/n,1/n)\).
Checkpoint 2. If \(x\ne0\), choose \(n\) large enough that \(1/n<|x|\); then \(x\notin(-1/n,1/n)\).
Checkpoint 3. Conclude the intersection is \(\{0\}\), then use the local interval criterion to show that singleton is not usual-open.
3. Learning check before the full solution.
The equality of the infinite intersection needs both inclusions. The Archimedean step 'choose \(n\) with \(1/n<|x|\)' excludes every nonzero point.
4. Complete step-by-step solution.
Proof. For every \(n\ge1\), \(0\in(-1/n,1/n)\), so \(0\) belongs to the intersection. Conversely, suppose \(x\ne0\). Since \(|x|>0\), the Archimedean property gives an integer \(n\) with \(n>1/|x|\), equivalently \(1/n<|x|\). Then \(x\notin(-1/n,1/n)\). Thus no nonzero real number belongs to every \(U_n\), and \[\bigcap_{n\ge1}(-1/n,1/n)=\{0\}.\] Each \(U_n\) is usual-open. However \(\{0\}\) is not usual-open: if it were, there would be \(\varepsilon>0\) with \((-\varepsilon,\varepsilon)\subseteq\{0\}\), but \(\varepsilon/2\) is a nonzero point of that interval. Therefore arbitrary intersections of open sets need not be open. \(\square\)
Expected evidence: Closure checks for finite unions and intersections, plus an explicit infinite union of members that leaves the collection.
1. Goal.
Let X=\(\mathbb N\) and let \(\mathcal C\) consist of \(\varnothing\), X, and all finite subsets of X. Show that \(\mathcal C\) is closed under finite unions and finite intersections but is not a topology.
2. Reasoning plan.
Checkpoint 1. Check finite unions and intersections by separating cases involving \(X\) from cases involving only finite sets.
Checkpoint 2. Choose infinitely many finite members whose union is infinite and still proper in \(\mathbb N\).
Checkpoint 3. The even numbers arise as the union of the singleton sets \(\{2n\}\).
3. Learning check before the full solution.
Closure under finite unions is not the topology axiom. A topology must be closed under arbitrary, possibly infinite, unions.
4. Complete step-by-step solution.
Solution. Let \(\mathcal C=\{\varnothing,\mathbb N\}\cup\{F\subseteq\mathbb N:F\text{ finite}\}\). A finite union of finite sets is finite; if one member is \(\mathbb N\), the union is \(\mathbb N\). Hence \(\mathcal C\) is closed under finite unions. A finite intersection of finite sets is finite; intersecting with \(\mathbb N\) changes nothing, so \(\mathcal C\) is also closed under finite intersections. Nevertheless, for each \(n\in\mathbb N\), the singleton \(\{2n\}\) is finite and therefore belongs to \(\mathcal C\). Their arbitrary union is \[\bigcup_{n\in\mathbb N}\{2n\}=\{0,2,4,\ldots\}.\] This set is infinite but is not all of \(\mathbb N\), so it belongs neither to the finite subsets nor to \(\{\mathbb N\}\). Therefore it is not in \(\mathcal C\), and arbitrary-union closure fails. Hence \(\mathcal C\) is not a topology.
Expected evidence: A pointwise local-ball proof for an arbitrary union of metric-open sets, with the chosen member and radius identified.
1. Goal.
Let (X,d) be a metric space. Prove directly that \(\tau_d\) is closed under arbitrary unions.
2. Reasoning plan.
Checkpoint 1. Let \(U=\bigcup_\alpha U_\alpha\) and choose an arbitrary \(x\in U\).
Checkpoint 2. Membership in a union gives one index \(\beta\) such that \(x\in U_\beta\).
Checkpoint 3. Use the radius witnessing openness of \(U_\beta\); the same ball lies in the larger union.
3. Learning check before the full solution.
A metric-open proof is pointwise: start with an arbitrary point of the candidate open set and construct a positive-radius ball around that point.
4. Complete step-by-step solution.
Proof. Let \((U_\alpha)_{\alpha\in A}\) be any family of sets in \(\tau_d\), and set \(U=\bigcup_{\alpha\in A}U_\alpha\). To prove \(U\in\tau_d\), take an arbitrary \(x\in U\). By definition of union, there exists an index \(\beta\in A\) such that \(x\in U_\beta\). Since \(U_\beta\) is metric-open, there is a radius \(r>0\) such that \(B_d(x,r)\subseteq U_\beta\). But \(U_\beta\subseteq U\), so \[B_d(x,r)\subseteq U_\beta\subseteq U.\] Thus the required ball exists for this arbitrary point \(x\). Therefore every point of \(U\) has a ball contained in \(U\), so \(U\) is metric-open. This proves closure under arbitrary unions. \(\square\)
Expected evidence: A local-ball proof for two intersections and a justified extension to arbitrary finite intersections, including the empty finite intersection.
1. Goal.
Let (X,d) be a metric space. Prove directly that \(\tau_d\) is closed under finite intersections.
2. Reasoning plan.
Checkpoint 1. Start with two open sets and a point in their intersection.
Checkpoint 2. Choose one witnessing radius for each set and take the smaller radius.
Checkpoint 3. For finitely many sets, take the minimum of finitely many positive radii; mention the empty finite intersection separately.
3. Learning check before the full solution.
The minimum of finitely many positive numbers is positive. That is the quantitative fact allowing one ball to fit simultaneously inside all finitely many open sets.
4. Complete step-by-step solution.
Proof. Let \(U,V\in\tau_d\) and take \(x\in U\cap V\). Because \(U\) is metric-open, choose \(r>0\) with \(B_d(x,r)\subseteq U\). Because \(V\) is metric-open, choose \(s>0\) with \(B_d(x,s)\subseteq V\). Let \(\rho=\min\{r,s\}\). Then \(\rho>0\), and every point at distance less than \(\rho\) from \(x\) is at distance less than both \(r\) and \(s\). Hence \[B_d(x,\rho)\subseteq U\cap V.\] So \(U\cap V\) is metric-open. For \(U_1,\ldots,U_n\), choose radii \(r_1,\ldots,r_n>0\) at a point \(x\) of the intersection and take \(\rho=\min_j r_j>0\). The same argument gives a ball inside all \(U_j\). The empty finite intersection is \(X\), which is open. Thus \(\tau_d\) is closed under finite intersections. \(\square\)
Expected evidence: Computation of a small discrete-metric ball and an argument showing that every subset satisfies the metric-open condition.
1. Goal.
Show that the discrete metric d(x,y)=0 if x=y and 1 otherwise induces the discrete topology.
2. Reasoning plan.
Checkpoint 1. For radius \(1/2\), compute exactly which points have discrete-metric distance less than \(1/2\) from \(x\).
Checkpoint 2. The result is the singleton \(\{x\}\).
Checkpoint 3. Given any subset \(A\) and any \(x\in A\), use that singleton ball as the local witness for openness.
3. Learning check before the full solution.
To prove the induced topology is discrete, prove every subset is metric-open. Showing only that singletons are open is enough only after using arbitrary unions, or directly via the local-ball criterion.
4. Complete step-by-step solution.
Proof. For the discrete metric, \(d(x,y)=0\) when \(y=x\) and \(d(x,y)=1\) when \(y\ne x\). Therefore \[B_d(x,1/2)=\{y\in X:d(x,y)<1/2\}=\{x\},\] because only \(y=x\) has distance below \(1/2\). Now let \(A\subseteq X\) be arbitrary. If \(A=\varnothing\), it is open. If \(x\in A\), the ball \(B_d(x,1/2)=\{x\}\) is contained in \(A\). Thus every point of \(A\) has a positive-radius ball contained in \(A\), so \(A\) is metric-open. Since the choice of \(A\subseteq X\) was arbitrary, every subset is open and \(\tau_d=\mathcal P(X)\), the discrete topology. \(\square\)
Expected evidence: An exact relation between balls for \(d\) and \(cd\), used in both directions to prove equality of the induced topologies.
1. Goal.
If d is a metric and c>0, prove that d and cd induce the same topology.
2. Reasoning plan.
Checkpoint 1. Let \(d_c=cd\). Solve the inequality \(cd(x,y)<r\) for \(d(x,y)\).
Checkpoint 2. This gives the exact equality \(B_{d_c}(x,r)=B_d(x,r/c)\).
Checkpoint 3. Use the equality in both directions: a \(d\)-ball witness gives a \(d_c\)-ball witness after multiplying the radius by \(c\), and conversely.
3. Learning check before the full solution.
Equality of topologies requires two inclusions. The ball identity makes the two local notions of openness equivalent, which gives both inclusions at once.
4. Complete step-by-step solution.
Proof. Define \(d_c(x,y)=c\,d(x,y)\) with \(c>0\). For every \(r>0\), \[B_{d_c}(x,r)=\{y:c\,d(x,y)<r\}=\{y:d(x,y)<r/c\}=B_d(x,r/c).\] Suppose \(U\) is \(d\)-open and \(x\in U\). Choose \(\rho>0\) such that \(B_d(x,\rho)\subseteq U\). Set \(r=c\rho>0\). Then \(B_{d_c}(x,r)=B_d(x,\rho)\subseteq U\), so \(U\) is \(d_c\)-open. Conversely, if \(U\) is \(d_c\)-open and \(x\in U\), choose \(r>0\) with \(B_{d_c}(x,r)\subseteq U\). Then \(B_d(x,r/c)=B_{d_c}(x,r)\subseteq U\), so \(U\) is \(d\)-open. Hence the two collections of open sets are equal: \(\tau_d=\tau_{cd}\). \(\square\)
Expected evidence: For each neighborhood claim, exhibit the open witness required by the definition and verify the needed containment.
1. Goal.
Prove that the intersection of two neighborhoods of x is again a neighborhood of x, and every superset of a neighborhood of x is a neighborhood of x.
2. Reasoning plan.
Checkpoint 1. For each neighborhood \(N_i\), choose an open set \(U_i\) with \(x\in U_i\subseteq N_i\).
Checkpoint 2. Use finite-intersection closure to show \(U_1\cap U_2\) is an open witness inside \(N_1\cap N_2\).
Checkpoint 3. For a superset \(M\supseteq N\), reuse the very same open witness that made \(N\) a neighborhood.
3. Learning check before the full solution.
The neighborhood itself does not have to be open. Always keep track of the open subset that witnesses the neighborhood property.
4. Complete step-by-step solution.
Proof. Suppose \(N_1\) and \(N_2\) are neighborhoods of \(x\). By definition, there exist open sets \(U_1,U_2\) such that \(x\in U_1\subseteq N_1\) and \(x\in U_2\subseteq N_2\). Because topologies are closed under finite intersections, \(U_1\cap U_2\) is open. Moreover \(x\in U_1\cap U_2\), and \[U_1\cap U_2\subseteq N_1\cap N_2.\] Thus \(U_1\cap U_2\) is an open witness showing that \(N_1\cap N_2\) is a neighborhood of \(x\). Now suppose \(N\) is a neighborhood of \(x\) and \(N\subseteq M\subseteq X\). Choose open \(U\) with \(x\in U\subseteq N\). Then \(x\in U\subseteq N\subseteq M\), so the same open set \(U\) witnesses that \(M\) is a neighborhood of \(x\). \(\square\)
Expected evidence: Both directions of the equivalence: use \(U\) itself forward and express \(U\) as a union of witnessing open sets backward.
1. Goal.
Prove that \(U\subseteq X\) is open if and only if U is a neighborhood of each of its points.
2. Reasoning plan.
Checkpoint 1. Forward: if \(U\) is open and \(x\in U\), use \(U\) itself as the required open witness.
Checkpoint 2. Backward: for every \(x\in U\), choose an open witness \(V_x\) with \(x\in V_x\subseteq U\).
Checkpoint 3. Prove \(U=\bigcup_{x\in U}V_x\) by two inclusions, then use arbitrary-union closure.
3. Learning check before the full solution.
The converse uses one possibly different open witness for each point. The topology axiom on arbitrary unions is exactly what assembles these local witnesses into the whole set \(U\).
4. Complete step-by-step solution.
Proof. Suppose first that \(U\) is open. For every \(x\in U\), choose the open set \(U\) itself: \(x\in U\subseteq U\). Hence \(U\) is a neighborhood of each of its points. Conversely, suppose \(U\) is a neighborhood of each point \(x\in U\). For every such \(x\), choose an open set \(V_x\) with \(x\in V_x\subseteq U\). Since \(V_x\subseteq U\) for every \(x\), \(\bigcup_{x\in U}V_x\subseteq U\). On the other hand, each \(x\in U\) belongs to its own \(V_x\), so \(U\subseteq\bigcup_{x\in U}V_x\). Therefore \[U=\bigcup_{x\in U}V_x.\] The right-hand side is an arbitrary union of open sets, hence open. Thus \(U\) is open if and only if it is a neighborhood of each of its points. \(\square\)
Expected evidence: Use topology inclusion for open sets and then pass to complements carefully for closed sets.
1. Goal.
Suppose \(\tau_1\subseteq\tau_2\) are topologies on X. Prove that every \(\tau_1\)-open set is \(\tau_2\)-open and every \(\tau_1\)-closed set is \(\tau_2\)-closed.
2. Reasoning plan.
Checkpoint 1. The open-set statement is exactly the inclusion \(\tau_1\subseteq\tau_2\).
Checkpoint 2. If \(F\) is \(\tau_1\)-closed, translate that into \(X\setminus F\in\tau_1\).
Checkpoint 3. Use the topology inclusion on the complement, then translate back to closedness in \(\tau_2\).
3. Learning check before the full solution.
A finer topology has more open sets and, correspondingly, also more closed sets, because taking complements gives a bijection between open and closed subsets.
4. Complete step-by-step solution.
Proof. Let \(U\) be \(\tau_1\)-open. This means \(U\in\tau_1\). Since \(\tau_1\subseteq\tau_2\), we immediately have \(U\in\tau_2\), so \(U\) is \(\tau_2\)-open. Now let \(F\) be \(\tau_1\)-closed. By definition, its complement \(X\setminus F\) is \(\tau_1\)-open, so \(X\setminus F\in\tau_1\subseteq\tau_2\). Therefore \(X\setminus F\) is \(\tau_2\)-open. Applying the definition of closed set again, this says that \(F\) is \(\tau_2\)-closed. Thus every open and every closed set for the coarser topology \(\tau_1\) remains open or closed, respectively, in the finer topology \(\tau_2\). \(\square\)
Expected evidence: One concrete subset, with separate proofs of its clopen status in the discrete topology and its failure to be open and closed in the usual topology.
1. Goal.
Give a subset of \(\mathbb R\) that is both open and closed in the discrete topology, but neither open nor closed in the usual topology.
2. Reasoning plan.
Checkpoint 1. Choose \(A=[0,1)\); every subset is open and closed in the discrete topology.
Checkpoint 2. To show \(A\) is not usual-open, focus on the point \(0\in A\) and show every interval around it leaves \(A\).
Checkpoint 3. To show \(A\) is not usual-closed, show its complement is not usual-open at the point \(1\), or equivalently note that \(1\) is a limit point not contained in \(A\).
3. Learning check before the full solution.
The exercise compares the same subset under two different topologies. Always name which topology is being used before asserting 'open' or 'closed'.
4. Complete step-by-step solution.
Solution. Let \(A=[0,1)\subseteq\mathbb R\). In the discrete topology \(\mathcal P(\mathbb R)\), every subset is open, so \(A\) is open. Its complement is also a subset of \(\mathbb R\) and is therefore open; hence \(A\) is also closed. Now switch to the usual topology. The point \(0\) belongs to \(A\), but for every \(\varepsilon>0\), the interval \((-\varepsilon,\varepsilon)\) contains negative points, so it is not contained in \(A\). Thus \(A\) is not usual-open. Its complement is \[\mathbb R\setminus A=(-\infty,0)\cup[1,\infty).\] The point \(1\) lies in this complement, but every open interval about \(1\) contains points below \(1\) that lie in \(A\). Hence the complement is not usual-open, so \(A\) is not usual-closed. Therefore \(A\) has the required behavior.
Expected evidence: Both strict inclusions, each supported by an explicit subset witnessing strictness.
1. Goal.
On an infinite set X, prove that the cofinite topology lies strictly between the indiscrete and discrete topologies.
2. Reasoning plan.
Checkpoint 1. The inclusions \(\{\varnothing,X\}\subseteq\tau_{\mathrm{cof}}\subseteq\mathcal P(X)\) are automatic.
Checkpoint 2. For strictness on the left, use \(X\setminus\{x\}\), a proper nonempty cofinite-open set.
Checkpoint 3. For strictness on the right, use \(\{x\}\); its complement is infinite when \(X\) is infinite, so it is not cofinite-open.
3. Learning check before the full solution.
To prove an inclusion is strict, exhibit an element belonging to the larger family but not to the smaller one.
4. Complete step-by-step solution.
Proof. Every topology contains \(\varnothing\) and \(X\), so \(\{\varnothing,X\}\subseteq\tau_{\mathrm{cof}}\). Also every cofinite-open set is a subset of \(X\), so \(\tau_{\mathrm{cof}}\subseteq\mathcal P(X)\). Let \(x\in X\). Because \(X\) is infinite, \(X\setminus\{x\}\) is nonempty and is not all of \(X\). Its complement is the finite singleton \(\{x\}\), so \(X\setminus\{x\}\in\tau_{\mathrm{cof}}\). Hence the left inclusion is strict. For the right inclusion, the singleton \(\{x\}\) belongs to \(\mathcal P(X)\). Its complement \(X\setminus\{x\}\) is infinite, so \(\{x\}\notin\tau_{\mathrm{cof}}\). Therefore \[\{\varnothing,X\}\subsetneq\tau_{\mathrm{cof}}\subsetneq\mathcal P(X).\] Thus the cofinite topology lies strictly between the indiscrete and discrete topologies. \(\square\)
Expected evidence: Represent an arbitrary subset as a union of singleton open sets and invoke arbitrary-union closure.
1. Goal.
If a topology \(\tau\) on \(X\) contains every singleton \(\{x\}\), prove that \(\tau\) is discrete.
2. Reasoning plan.
Checkpoint 1. Fix an arbitrary subset \(A\subseteq X\).
Checkpoint 2. Write it as \(A=\bigcup_{x\in A}\{x\}\).
Checkpoint 3. Every singleton in this union is open by hypothesis, so arbitrary-union closure makes \(A\) open.
3. Learning check before the full solution.
For \(A=\varnothing\), the union over an empty index set is \(\varnothing\), which is open by the topology axiom. Thus the same formula covers every subset.
4. Complete step-by-step solution.
Proof. Assume that \(\{x\}\in\tau\) for every \(x\in X\). Let \(A\subseteq X\) be arbitrary. Every element of \(A\) lies in its singleton, and every singleton \(\{x\}\) with \(x\in A\) is contained in \(A\). Hence \[A=\bigcup_{x\in A}\{x\}.\] By hypothesis each \(\{x\}\) in this family is open. A topology is closed under arbitrary unions, so the right-hand side belongs to \(\tau\). Thus \(A\in\tau\). Since \(A\subseteq X\) was arbitrary, every subset of \(X\) is open, so \(\mathcal P(X)\subseteq\tau\). The reverse inclusion \(\tau\subseteq\mathcal P(X)\) holds by definition. Therefore \(\tau=\mathcal P(X)\), and the topology is discrete. \(\square\)
Expected evidence: Construct each singleton as a finite intersection of the given co-singleton open sets, handle the one-point edge case, then conclude discreteness.
1. Goal.
Let X be finite. If \(X\setminus\{x\}\) is open for every \(x\in X\), prove the topology is discrete.
2. Reasoning plan.
Checkpoint 1. Fix \(x\in X\) and intersect all sets \(X\setminus\{y\}\) with \(y\ne x\).
Checkpoint 2. Because \(X\) is finite, only finitely many sets occur, so the intersection is open.
Checkpoint 3. Show the intersection equals \(\{x\}\); then invoke the singleton criterion from Exercise 3.25. Handle \(|X|=1\) using the empty finite intersection \(X\).
3. Learning check before the full solution.
The finiteness of \(X\) is essential because a topology is guaranteed to be closed only under finite intersections of open sets, not arbitrary intersections.
4. Complete step-by-step solution.
Proof. Fix \(x\in X\). For every \(y\in X\setminus\{x\}\), the set \(X\setminus\{y\}\) is open by hypothesis. Since \(X\) is finite, there are only finitely many such \(y\), so \[V_x=\bigcap_{\substack{y\in X\\y\ne x}}(X\setminus\{y\})\] is open. A point \(z\in X\) belongs to \(V_x\) exactly when \(z\ne y\) for every \(y\ne x\). The only possible such point is \(z=x\), and \(x\) indeed belongs to every factor. Hence \(V_x=\{x\}\). If \(X=\{x\}\), the displayed intersection is the empty finite intersection, equal to \(X=\{x\}\), still open. Thus every singleton is open. By Exercise 3.25, every subset is a union of open singletons, so every subset is open. Therefore the topology is discrete. \(\square\)
Expected evidence: Exhibit a topology containing the two required sets, verify it, and prove minimality by showing every admissible topology must contain all four members.
1. Goal.
For \(X=\{a,b,c\}\), determine the smallest topology containing {a} and {b,c}.
2. Reasoning plan.
Checkpoint 1. Any topology containing the required sets must also contain \(\varnothing\) and \(X\).
Checkpoint 2. Check whether \(\{\varnothing,\{a\},\{b,c\},X\}\) is already closed under arbitrary unions and finite intersections.
Checkpoint 3. For minimality, let \(\sigma\) be any topology containing \(\{a\}\) and \(\{b,c\}\) and show all four displayed sets lie in \(\sigma\).
3. Learning check before the full solution.
'Smallest topology containing' means two things: your candidate is a topology with the required sets, and it is contained in every other topology having those sets.
4. Complete step-by-step solution.
Solution. Consider \(\tau=\{\varnothing,\{a\},\{b,c\},X\}\). It contains \(\varnothing\) and \(X\). The two nontrivial sets are disjoint, with \(\{a\}\cap\{b,c\}=\varnothing\), and their union is \(X\). Thus every union or finite intersection of members of \(\tau\) is again one of the four displayed sets; hence \(\tau\) is a topology. It contains the required sets \(\{a\}\) and \(\{b,c\}\). To prove minimality, let \(\sigma\) be any topology on \(X\) containing those two sets. Axiom (T1) forces \(\varnothing,X\in\sigma\), while the requirement gives \(\{a\},\{b,c\}\in\sigma\). Therefore every member of \(\tau\) belongs to \(\sigma\), so \(\tau\subseteq\sigma\). Hence \(\tau\) is the smallest topology containing the two specified subsets. \(\square\)
Expected evidence: Check each candidate separately, identifying the exact failed axiom for the non-topology and verifying all nontrivial operations for the valid ones.
1. Goal.
For \(X=\{a,b,c\}\), decide which of the following are topologies: \(A=\{\varnothing,\{a\},X\}\); \(B=\{\varnothing,\{a\},\{b\},X\}\); \(C=\{\varnothing,\{a\},\{b\},\{a,b\},X\}\).
2. Reasoning plan.
Checkpoint 1. For \(A\), the members are nested, which simplifies unions and intersections.
Checkpoint 2. For \(B\), test \(\{a\}\cup\{b\}=\{a,b\}\); the missing union is enough to fail (T2).
Checkpoint 3. For \(C\), verify the only genuinely new union and the intersections of the proper nonempty members.
3. Learning check before the full solution.
When a candidate fails one topology axiom, stop: one explicit failure is sufficient. For candidates that pass, you must still justify all three axioms.
4. Complete step-by-step solution.
Solution. For \(A=\{\varnothing,\{a\},X\}\), (T1) holds and the members are nested, so arbitrary unions and finite intersections remain in \(A\). Thus \(A\) is a topology. For \(B=\{\varnothing,\{a\},\{b\},X\}\), the sets \(\{a\}\) and \(\{b\}\) are members, but their union \(\{a,b\}\) is not. Therefore \(B\) fails (T2) and is not a topology. For \(C=\{\varnothing,\{a\},\{b\},\{a,b\},X\}\), (T1) holds. The union \(\{a\}\cup\{b\}=\{a,b\}\) is present; unions involving \(\{a,b\}\) remain \(\{a,b\}\) or become \(X\), and arbitrary unions reduce to these possibilities. Intersections satisfy \(\{a\}\cap\{b\}=\varnothing\), while intersections with \(\{a,b\}\) give \(\{a\}\) or \(\{b\}\). Thus (T2) and (T3) hold, so \(C\) is a topology.
Expected evidence: First use the intersection theorem, then prove the universal 'largest common subtopology' property by two containments.
1. Goal.
Let \(\tau_1\) and \(\tau_2\) be topologies on X. Explain why \(\tau_1\cap\tau_2\) is the largest topology contained in both \(\tau_1\) and \(\tau_2\).
2. Reasoning plan.
Checkpoint 1. Theorem 3.2 gives that \(\tau_1\cap\tau_2\) is itself a topology.
Checkpoint 2. By definition of set intersection, it is contained in both \(\tau_1\) and \(\tau_2\).
Checkpoint 3. If \(\sigma\) is any topology contained in both, each \(U\in\sigma\) belongs to both, hence to their intersection.
3. Learning check before the full solution.
'Largest' is with respect to inclusion among topologies. It does not mean largest number of points or largest open set.
4. Complete step-by-step solution.
Proof. By Theorem 3.2, \(\tau_1\cap\tau_2\) is a topology on \(X\). Because it is a set-theoretic intersection, \(\tau_1\cap\tau_2\subseteq\tau_1\) and \(\tau_1\cap\tau_2\subseteq\tau_2\). Thus it is a topology contained in both. Now let \(\sigma\) be any topology such that \(\sigma\subseteq\tau_1\) and \(\sigma\subseteq\tau_2\). Take \(U\in\sigma\). The first inclusion gives \(U\in\tau_1\), and the second gives \(U\in\tau_2\). Therefore \(U\in\tau_1\cap\tau_2\). Since \(U\) was arbitrary, \(\sigma\subseteq\tau_1\cap\tau_2\). Hence every topology contained in both \(\tau_1\) and \(\tau_2\) is contained in their intersection. This is exactly the universal property of being the largest common subtopology. \(\square\)
Expected evidence: A strict inclusion chain, the corresponding finer/coarser interpretation, and a complete complement computation for the closed sets of \(\tau_2\).
1. Goal.
On \(X=\{a,b,c\}\), compare \(\tau_0=\{\varnothing,X\}\), \(\tau_1=\{\varnothing,\{a\},X\}\), \(\tau_2=\{\varnothing,\{a\},\{a,b\},X\}\), and \(\tau_3=\mathcal P(X)\) by inclusion, and list the closed sets of \(\tau_2\).
2. Reasoning plan.
Checkpoint 1. Compare successive lists of open sets to obtain the strict inclusion chain.
Checkpoint 2. Translate the direction of inclusion into increasing fineness.
Checkpoint 3. For \(\tau_2\), take complements in \(X=\{a,b,c\}\) of each of its four open sets.
3. Learning check before the full solution.
The closed sets of \(\tau_2\) are determined by \(\tau_2\), not by the larger topology \(\tau_3\). Keep the topology fixed while taking complements.
4. Complete step-by-step solution.
Solution. We have \(\tau_0=\{\varnothing,X\}\). The topology \(\tau_1\) adds the open set \(\{a\}\), so \(\tau_0\subsetneq\tau_1\). The topology \(\tau_2\) adds \(\{a,b\}\), so \(\tau_1\subsetneq\tau_2\). Finally \(\tau_3=\mathcal P(X)\) contains all subsets, including for example \(\{b\}\notin\tau_2\), so \(\tau_2\subsetneq\tau_3\). Hence \[\tau_0\subsetneq\tau_1\subsetneq\tau_2\subsetneq\tau_3.\] Fineness increases from left to right. To list the \(\tau_2\)-closed sets, take complements of its open sets: \(X\setminus\varnothing=X\), \(X\setminus\{a\}=\{b,c\}\), \(X\setminus\{a,b\}=\{c\}\), and \(X\setminus X=\varnothing\). Thus the closed sets are exactly \(X,\{b,c\},\{c\},\varnothing\).
Chapter 3 extracts the structural content hidden inside metric openness. A topology specifies which subsets count as open, closed sets arise by complementation, neighborhoods localize the structure at a point, and inclusion of topologies records how much openness a space is being given.
Chapter 4 now asks how such a topology can be described economically from smaller generating families. That leads to bases and subbases without changing the topology axioms established here.