Exercise Set 2 (Homework 03 & 04)
This section contains exercises related to the third and fourth homework assignments of the Point-Set Topology course.
Problem 1
Let \(S\) be a non-empty set.
(i) Consider the discrete metric \(d: S \times S \to \mathbb{R}_{\ge 0}\), defined as: \(d(x, y)=0\) when \(x=y\); and \(d(x, y)=1\) when \(x \neq y\). Prove that the topology \(\mathcal{T}_d\) induced by this metric is exactly the power set \(\mathcal{P}(S)\), i.e., every subset of \(S\) is an open set in the metric space \((S, d)\).
(ii) Is it always possible to find a metric on \(S\) such that the topology it induces is the trivial topology (i.e., \(\{ \emptyset, S \}\))?
Solution
(i) Prove that the discrete metric induces the discrete topology
To prove \(\mathcal{T}_d = \mathcal{P}(S)\), it is sufficient to prove that every singleton set \(\{x\}\) in \(S\) is an open set.
For any \(x \in S\), consider the open ball centered at \(x\) with radius \(\epsilon = 1/2\):
According to the definition of the discrete metric, when \(y \neq x\), \(d(y, x) = 1 > 1/2\); only when \(y = x\), \(d(y, x) = 0 < 1/2\). Therefore, \(B(x, 1/2) = \{x\}\).
Since an open ball in a metric space must be an open set, \(\{x\}\) is an open set. Because any subset \(A\) of \(S\) can be expressed as the union of singleton sets:
and the topology is closed under arbitrary union operations, \(A\) must be an open set. Thus, \(\mathcal{T}_d = \mathcal{P}(S)\).
(ii) Discussion on the metrizability of the trivial topology
Conclusion: It is possible when \(|S| \le 1\); it does not exist when \(|S| \ge 2\).
Reasoning: Assume \(S\) contains at least two distinct points \(x\) and \(y\). If there exists a metric \(d\) inducing the trivial topology, then \(d(x, y) = \epsilon > 0\). Consider the open ball \(B(x, \epsilon) = \{ z \in S \mid d(z, x) < \epsilon \}\).
Since \(d(x, x) = 0 < \epsilon\), then \(x \in B(x, \epsilon)\). Since \(d(y, x) = \epsilon \not< \epsilon\), then \(y \notin B(x, \epsilon)\).
This means \(B(x, \epsilon)\) is neither the empty set (it contains \(x\)) nor the entire space \(S\) (it does not contain \(y\)). However, in the trivial topology, the only open sets are \(\emptyset\) and \(S\). This creates a contradiction. Therefore, if \(|S| \ge 2\), there is no metric that induces the trivial topology.
Problem 2
Prove that \(\mathcal{B} := \{ (a, b) \mid a < b \text{ and } a, b \in \mathbb{Q} \}\) is a basis for the standard topology on \(\mathbb{R}\) (induced by the metric \(d(x, y) = |x - y|\)). Also prove that \(\text{Card}(\mathcal{B}) = \aleph_0\).
Solution
1. Prove that \(\mathcal{B}\) is a basis for the standard topology
Let \(\mathcal{T}\) be the standard topology on \(\mathbb{R}\). To prove that \(\mathcal{B}\) is a basis, we need to prove that for any open set \(U \in \mathcal{T}\) and any point \(x \in U\), there exists \(B \in \mathcal{B}\) such that \(x \in B \subseteq U\).
Given that \(U\) is an open set, there exists \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon) \subseteq U\). According to the Density of rational numbers in real numbers: In the interval \((x - \epsilon, x)\), there must exist a rational number \(a\); In the interval \((x, x + \epsilon)\), there must exist a rational number \(b\).
Thus we obtain \(a < x < b\), and \(a, b \in \mathbb{Q}\). Clearly \((a, b) \in \mathcal{B}\). And because \(x - \epsilon < a\) and \(b < x + \epsilon\), we have:
Proof complete.
2. Calculate the cardinality of the basis
Each element of the set \(\mathcal{B}\) is uniquely determined by an ordered pair of rational numbers \((a, b)\) satisfying \(a < b\). Therefore, there exists an injection \(\Psi: \mathcal{B} \to \mathbb{Q} \times \mathbb{Q}\). Since \(\mathbb{Q}\) is a countable set, its Cartesian product \(\mathbb{Q} \times \mathbb{Q}\) is also a countable set (i.e., its cardinality is \(\aleph_0\)). Because \(\mathcal{B}\) is an infinite set (e.g., containing all \((0, n), n \in \mathbb{Q}^+\)), and its cardinality is not greater than \(\aleph_0\), by the Cantor-Bernstein Theorem:
Problem 3
For \(n \in \mathbb{N}_{\ge 1}\), consider a finite set \(X_n := \{1, 2, \dots, n\}\) of size \(n\). How many topologies are there on \(X_n\)? Furthermore, consider the set \(\Omega\) of all topologies on \(X_n\). Define \(\mathcal{T}_i < \mathcal{T}_j\) as \(\mathcal{T}_i \subsetneq \mathcal{T}_j\). Under what conditions does \((\Omega, <)\) form a linear order relation?
Solution
1. Number of Topologies
There is no simple general formula for the number of topologies on a finite set, usually denoted as \(T(n)\). For the first few \(n\), the values are:
- \(n=1: T(1) = 1\) (only the trivial topology)
- \(n=2: T(2) = 4\) (trivial, discrete, and two Sierpinski topologies)
- \(n=3: T(3) = 29\)
2. Discussion on the Linear Order Relation
Conclusion: \((\Omega, <)\) forms a linear order relation if and only if \(n=1\).
Reasoning: A linear order requires that for any two elements \(\mathcal{T}_i, \mathcal{T}_j\) in \(\Omega\), it must hold that \(\mathcal{T}_i \subseteq \mathcal{T}_j\) or \(\mathcal{T}_j \subseteq \mathcal{T}_i\).
When \(n \ge 2\), we can construct two incomparable topologies. Suppose \(X_n\) contains distinct elements \(a\) and \(b\). Define topology \(\mathcal{T}_1 = \{ \emptyset, X_n, \{a\} \}\); Define topology \(\mathcal{T}_2 = \{ \emptyset, X_n, \{b\} \}\).
Clearly \(\{a\} \notin \mathcal{T}_2\) and \(\{b\} \notin \mathcal{T}_1\). Thus \(\mathcal{T}_1 \not\subseteq \mathcal{T}_2\) and \(\mathcal{T}_2 \not\subseteq \mathcal{T}_1\). Therefore, when \(n \ge 2\), \((\Omega, <)\) is not a linear order.
Problem 4
Let \(S\) be a set.
(i) Prove: If \(\{ \mathcal{T}_i \}_{i \in I}\) is a family of topologies on \(S\), then their intersection \(\bigcap_{i \in I} \mathcal{T}_i\) is also a topology on \(S\). (ii) Give an example of two topologies \(\mathcal{T}_1\) and \(\mathcal{T}_2\) such that \(\mathcal{T}_1 \cup \mathcal{T}_2\) is not a topology. Find the unique smallest topology containing \(\mathcal{T}_1 \cup \mathcal{T}_2\).
Solution
(i) Prove that the intersection is a topology
Denote \(\mathcal{T} = \bigcap_{i \in I} \mathcal{T}_i\).
- Contains the empty set and the whole space: Since for all \(i \in I\), \(\emptyset, S \in \mathcal{T}_i\) holds, then \(\emptyset, S \in \mathcal{T}\).
- Closed under arbitrary unions: Let \(\{ U_\alpha \}_{\alpha \in A} \subseteq \mathcal{T}\). Then for each \(i \in I\), \(\{ U_\alpha \} \subseteq \mathcal{T}_i\). Since \(\mathcal{T}_i\) is a topology, \(\bigcup U_\alpha \in \mathcal{T}_i\). Therefore \(\bigcup U_\alpha \in \bigcap \mathcal{T}_i = \mathcal{T}\).
- Closed under finite intersections: Let \(U_1, \dots, U_k \in \mathcal{T}\). Then for each \(i \in I\), \(U_1, \dots, U_k \in \mathcal{T}_i\), so \(\bigcap_{j=1}^k U_j \in \mathcal{T}_i\). Therefore, the finite intersection belongs to \(\mathcal{T}\).
(ii) Counterexample where the union is not a topology
Let \(S = \{a, b, c\}\). Take \(\mathcal{T}_1 = \{ \emptyset, S, \{a\} \}\) and \(\mathcal{T}_2 = \{ \emptyset, S, \{b\} \}\). Then \(\mathcal{T}_1 \cup \mathcal{T}_2 = \{ \emptyset, S, \{a\}, \{b\} \}\).
However, \(\{a\} \cup \{b\} = \{a, b\}\), but \(\{a, b\} \notin \mathcal{T}_1 \cup \mathcal{T}_2\). This violates the axiom of closure under union operations, so the union is not a topology.
Smallest Topology: The unique smallest topology containing \(\mathcal{T}_1 \cup \mathcal{T}_2\) is the topology generated by this union as a subbasis. In this example, the topology is:
Problem 5
A subset of \(\mathbb{R}^2\) is called radially open if it contains an open line segment in every direction at each of its points. Let \(\mathcal{T}_{ro}\) denote the family of all radially open sets.
Prove that \(\mathcal{T}_{ro}\) is a topology on \(\mathbb{R}^2\). What is its relationship with the standard topology \(\mathcal{T}_{std}\)?
Solution
1. Prove it is a topology
The key lies in proving closure under finite intersections. Let \(A, B \in \mathcal{T}_{ro}\). For any point \(x \in A \cap B\):
- Since \(x \in A\), there exists an open line segment \(I_A(\theta) \subseteq A\) in direction \(\theta\);
- Since \(x \in B\), there exists an open line segment \(I_B(\theta) \subseteq B\) in direction \(\theta\).
Take the intersection of these two segments \(I_A \cap I_B\). Since they share the same center and are on the same ray, their intersection is still an open line segment containing \(x\), and \(I_A \cap I_B \subseteq A \cap B\). Therefore \(A \cap B\) is also radially open. Combining with the obvious inclusion of \(\emptyset, S\) and the properties of union operations, \(\mathcal{T}_{ro}\) is a topology.
2. Relationship with the standard topology
Conclusion: \(\mathcal{T}_{std} \subsetneq \mathcal{T}_{ro}\).
- Inclusion relation: Points in a standard open set contain an open ball \(B(x, \epsilon)\). This ball clearly contains an open line segment of length \(2\epsilon\) in any direction. Thus, a standard open set must be a radially open set.
- Strictness (Counterexample): There exist radially open sets that are not standard open sets. Construct a set: \(U = \{ (r\cos\theta, r\sin\theta) \mid 0 \le r < f(\theta) \}\), where \(f(\theta)\) is a positive function that approaches 0 rapidly as the polar angle approaches certain values. For example, if \(f(\theta)\) is made extremely "sharp" near the origin, the origin contains segments in every direction but cannot contain any full disk. Such a set is radially open at the origin, but the origin is not an interior point under the standard topology.
Problem 6
Let \(X = \mathbb{R}^2\) endowed with the standard topology. Consider its subset (Topologist's Sine Curve):
Calculate the following four sets: \(A^o\), \(LP(A)\), \(\overline{A}\), and \(\overline{A} \cap \overline{\mathbb{R}^2 - A}\).
Solution
1. Interior \(A^o\)
Since \(A\) is the graph of the continuous function \(y = \sin(1/x)\) on the interval \((0, 1]\), it is a curve in \(\mathbb{R}^2\). For any point \(P\) in \(A\), any of its open neighborhoods (open disks) will contain points not belonging to the curve (i.e., points whose y-coordinates do not satisfy \(y = \sin(1/x)\)). Therefore, \(A\) does not contain any non-empty open balls. Thus:
2. Limit Point Set \(LP(A)\) and Closure \(\overline{A}\)
As \(x \to 0^+\), the function \(\sin(1/x)\) oscillates infinitely many times between \(-1\) and \(1\). Therefore, every point in the segment \(\{0\} \times [-1, 1]\) on the y-axis is a limit point of \(A\). The points of set \(A\) itself are obviously also its limit points. Therefore:
Since the closure \(\overline{A} = A \cup LP(A)\), we can conclude:
3. Intersection \(\overline{A} \cap \overline{\mathbb{R}^2 - A}\)
Since \(A^o = \emptyset\), every point in the whole space belongs either to the interior of \(\mathbb{R}^2 - A\) or to the boundary of \(A\). Since \(\overline{\mathbb{R}^2 - A} = \mathbb{R}^2 - A^o = \mathbb{R}^2 - \emptyset = \mathbb{R}^2\). Therefore:
Problem 7
Let \(X\) be a set, and \(f: \mathcal{P}(X) \to \mathcal{P}(X)\) be a function satisfying the following properties (for any \(A, B \subseteq X\)): (i) \(f(\emptyset) = \emptyset\); (ii) \(A \subseteq f(A)\); (iii) \(f(A) = f(f(A))\); (iv) \(f(A \cup B) = f(A) \cup f(B)\).
Prove: (1) The set \(\mathcal{T}_f := \{ X - A : A \in \mathcal{P}(X) \text{ and } f(A) = A \}\) is a topology on \(X\); (2) The closure \(\overline{A}\) of any subset \(A \subseteq X\) with respect to the topology \(\mathcal{T}_f\) is exactly \(f(A)\).
Solution
(1) Prove that \(\mathcal{T}_f\) is a topology
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Empty set and Whole space: From property (i), \(f(\emptyset) = \emptyset\), thus \(X - \emptyset = X \in \mathcal{T}_f\). From property (ii), \(X \subseteq f(X)\), while \(f(X) \subseteq X\) obviously holds, so \(f(X) = X\). Therefore \(X - X = \emptyset \in \mathcal{T}_f\).
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Arbitrary Union Operation: Let \(\{ U_\alpha \}_{\alpha \in I}\) be a family of open sets in \(\mathcal{T}_f\), where \(U_\alpha = X - A_\alpha\) and \(f(A_\alpha) = A_\alpha\). Their union \(\bigcup U_\alpha = X - \bigcap A_\alpha\). From property (iv), it can be deduced that \(f\) is monotonic: if \(A \subseteq B\), then \(f(A) \subseteq f(B)\). Since \(\bigcap A_\alpha \subseteq A_\alpha\), then \(f(\bigcap A_\alpha) \subseteq f(A_\alpha) = A_\alpha\). Thus \(f(\bigcap A_\alpha) \subseteq \bigcap A_\alpha\). Combining with the reverse inclusion in property (ii), we get \(f(\bigcap A_\alpha) = \bigcap A_\alpha\). Therefore \(\bigcup U_\alpha \in \mathcal{T}_f\).
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Finite Intersection Operation: Let \(U_1, U_2 \in \mathcal{T}_f\), then \(U_1 \cap U_2 = X - (A_1 \cup A_2)\). From property (iv), \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2) = A_1 \cup A_2\). Therefore \(U_1 \cap U_2 \in \mathcal{T}_f\).
(2) Prove the closure \(\overline{A} = f(A)\)
According to the definition of closure, \(\overline{A}\) is the intersection of all closed sets containing \(A\). Under this topology, a closed set \(C\) satisfies \(f(C) = C\).
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First, from property (iii), \(f(f(A)) = f(A)\), so \(f(A)\) itself is a closed set. Also, from property (ii), \(A \subseteq f(A)\). Thus \(f(A)\) is a closed set containing \(A\), so \(\overline{A} \subseteq f(A)\).
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Second, let \(C\) be any closed set containing \(A\) (i.e., \(A \subseteq C\) and \(f(C) = C\)). From the monotonicity of \(f\), we have \(f(A) \subseteq f(C) = C\). This means \(f(A)\) is contained in every closed set containing \(A\), therefore \(f(A) \subseteq \bigcap C = \overline{A}\).
In summary, \(\overline{A} = f(A)\).
Problem 8
Consider the function \(f: [0, 1) \to S^1\), defined as \(f(t) = (\cos(2\pi t), \sin(2\pi t))\). Prove that \(f\) is not a homeomorphism with respect to the standard subspace topology. Is it possible to find a homeomorphism between \([0, 1)\) and \(S^1\)?
Solution
1. Prove \(f\) is not a homeomorphism
Although \(f\) is a continuous bijection, its inverse \(f^{-1}\) is discontinuous at point \((1, 0)\). We can provide a more rigorous proof from the perspective of topological invariants:
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Method 1: Compactness Under the standard Euclidean topology, \(S^1\) is a bounded closed set in \(\mathbb{R}^2\), and therefore it is compact. However, \([0, 1)\) is not a closed set in \(\mathbb{R}\), and therefore it is not compact. Since a homeomorphism must preserve compactness, \(f\) cannot be a homeomorphism.
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Method 2: Cut point (Connectedness) If \(f\) is a homeomorphism, then the connectedness of the space after removing a point should remain unchanged. For \(S^1\), removing any point \(p\) leaves the remaining space \(S^1 - \{p\}\) still connected (homeomorphic to an open interval). For \([0, 1)\), if we remove point \(t = 1/2\), the remaining space \([0, 1/2) \cup (1/2, 1)\) is disconnected. Therefore, these two spaces do not have the same local connected structure, so \(f\) is not a homeomorphism.
2. Conclusion
Impossible. Based on the reasons of compactness and cut points above, no homeomorphism exists between \([0, 1)\) and \(S^1\).
Problem 9
A subset \(A \subseteq X\) is called regularly open if \(A = (\overline{A})^o\).
(i) Give an example of an open set in \(\mathbb{R}\) that is not regularly open. Can you characterize regularly open sets in \(\mathbb{R}\)? (ii) Prove that for any \(A \subseteq X\), the set \((\overline{A})^o\) is a regularly open set.
Solution
(i) Example and Characterization
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Counterexample: Take \(A = (0, 1) \cup (1, 2)\). Then \(\overline{A} = [0, 2]\), and \((\overline{A})^o = (0, 2)\). Clearly \(A \neq (\overline{A})^o\), so \(A\) is not a regularly open set. The reason is that \(A\) is missing a point \(\{1\}\) from the interior of the closure.
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Characterization: Regularly open sets in \(\mathbb{R}\) are those open sets "without extra gaps". That is, they cannot be obtained by removing some isolated points or boundary points from a larger open set.
(ii) Prove \((\overline{A})^o\) is regularly open
Let \(U = (\overline{A})^o\). We want to prove \(U = (\overline{U})^o\).
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According to the definition \(U \subseteq \overline{U}\), since \(U\) is an open set, \(U = U^o \subseteq (\overline{U})^o\).
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On the other hand, since \(U = (\overline{A})^o \subseteq \overline{A}\), and the closure operation is monotonic and idempotent: It follows that \(\overline{U} \subseteq \overline{(\overline{A})} = \overline{A}\). Taking the interior of both sides gives \((\overline{U})^o \subseteq (\overline{A})^o = U\).
In summary, \(U = (\overline{U})^o\), i.e., \((\overline{A})^o\) is a regularly open set.
Problem 10
Let \(A\) be a subset of a topological space \(X\). Prove the equation \(\overline{A} = A \cup LP(A)\).
Solution
1. Prove \(A \cup LP(A) \subseteq \overline{A}\)
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It is obvious that \(A \subseteq \overline{A}\).
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Let \(x \in LP(A)\). According to the definition of limit point, for any open neighborhood \(U\) of \(x\), \(U \cap (A - \{x\}) \neq \emptyset\). This means \(U \cap A \neq \emptyset\) holds for all open neighborhoods. According to the characterization of closure by neighborhoods, this is exactly the necessary and sufficient condition for \(x \in \overline{A}\). Thus \(LP(A) \subseteq \overline{A}\).
Therefore \(A \cup LP(A) \subseteq \overline{A}\).
2. Prove \(\overline{A} \subseteq A \cup LP(A)\)
Let \(x \in \overline{A}\).
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If \(x \in A\), then the conclusion obviously holds.
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If \(x \notin A\), since \(x \in \overline{A}\), for any open neighborhood \(U\) of \(x\), \(U \cap A \neq \emptyset\). Because \(x \notin A\), the points in the set \(U \cap A\) must not be \(x\). That is, \(U \cap (A - \{x\}) \neq \emptyset\). This satisfies the definition of a limit point, so \(x \in LP(A)\).
Therefore \(x \in A \cup LP(A)\).
In summary, \(\overline{A} = A \cup LP(A)\).