Exercise Set 3 (Homework 05 & 06)
This section contains the fifth and sixth assignments for the Point Set Topology course and related supplementary exercises. Topics include the equivalence of metric spaces, infinite product spaces, connectivity and path-connectivity, fixed-point theorems, and the compactness of the finite-complement topology.
Problem 1
Let \((X, d)\) be a metric space, and let \(\mathcal{T}\) be the topology induced by \(d\). Define a function:
for every pair \(x, y \in X\).
Prove:
(i) \(d^*\) is a metric on \(X\). (ii) The topology \(\mathcal{T}^*\) induced by \(d^*\) is equal to \(\mathcal{T}\). (iii) Is it always possible to find a constant \(A \in \mathbb{R}_{>0}\) such that for all \(x, y \in X\), \(A \cdot d(x, y) \le d^*(x, y)\)?
Solution
(i) Proving \(d^*\) is a metric
-
Non-negativity and Positivity: Since \(d(x, y) \ge 0\), then \(d^*(x, y) = \min\{d, 1\} \ge 0\). Furthermore, \(d^*(x, y) = 0 \iff \min\{d, 1\} = 0 \iff d(x, y) = 0 \iff x = y\).
-
Symmetry: \(d^*(x, y) = \min\{d(x, y), 1\} = \min\{d(y, x), 1\} = d^*(y, x)\).
-
Triangle Inequality: We need to prove \(d^*(x, z) \le d^*(x, y) + d^*(y, z)\).
-
If \(d^*(x, y) + d^*(y, z) \ge 1\), since \(d^*(x, z) \le 1\), the inequality holds trivially.
- If \(d^*(x, y) + d^*(y, z) < 1\), since \(d^* \le d\), this implies \(d^*(x, y) = d(x, y)\) and \(d^*(y, z) = d(y, z)\). According to the triangle inequality of the original metric \(d\): \(d(x, z) \le d(x, y) + d(y, z) < 1\). Therefore, \(d^*(x, z) = \min\{d(x, z), 1\} = d(x, z)\). Thus, \(d^*(x, z) \le d(x, y) + d(y, z) = d^*(x, y) + d^*(y, z)\).
(ii) Proving \(\mathcal{T}^* = \mathcal{T}\)
To prove that two metrics induce the same topology, it suffices to show that their open balls can contain each other.
-
For any \(\epsilon > 0\), let \(\epsilon' = \min\{\epsilon, 1\}\). Then \(B_{d^*}(x, \epsilon') \subseteq B_d(x, \epsilon)\). Because if \(d^*(x, y) < \epsilon' \le 1\), then \(d(x, y) = d^*(x, y) < \epsilon' \le \epsilon\).
-
Conversely, for any \(\epsilon > 0\), let \(\epsilon' = \min\{\epsilon, 1/2\}\). Then \(B_d(x, \epsilon') \subseteq B_{d^*}(x, \epsilon)\). Because if \(d(x, y) < \epsilon' \le 1/2\), then \(d^*(x, y) = d(x, y) < \epsilon' \le \epsilon\).
Since the open balls contain each other, the topologies generated by the two metrics are identical.
(iii) Discussion on Constant \(A\)
Conclusion: Not necessarily.
Reason: If the metric \(d\) is unbounded (for example, in Euclidean space \(\mathbb{R}^n\)), for any \(A > 0\), we can always find two points \(x, y\) far enough apart such that \(d(x, y) > 1/A\). In this case:
However, by definition \(d^*(x, y) \le 1\) always holds. This leads to \(A \cdot d(x, y) > d^*(x, y)\), which contradicts the requirement. This \(A\) can only be found if the original metric space itself is bounded.
Problem 2
Consider the Cartesian product space:
Let \(R_{ez}^\omega \subseteq \mathbb{R}^\omega\) be the subset of all "eventually zero" sequences (i.e., only finitely many components are non-zero).
Please calculate the interior and closure of \(R_{ez}^\omega\) under the Product Topology \(\mathcal{T}_p\) and the Box Topology \(\mathcal{T}_b\), respectively.
Solution
1. Under the Product Topology \(\mathcal{T}_p\)
-
Interior \((R_{ez}^\omega)^\circ\): Any non-empty open set \(U\) must contain a basis element \(\prod U_i\) (where almost all \(U_i = \mathbb{R}\)). Since infinitely many components can take any real value, this open set must contain sequences with infinitely many non-zero components. Therefore, \(U \not\subseteq R_{ez}^\omega\), so the interior is empty: \((R_{ez}^\omega)^\circ = \emptyset\).
-
Closure \(\overline{R_{ez}^\omega}\): For any sequence \(x = (x_i) \in \mathbb{R}^\omega\) and any open neighborhood \(U\), we can choose a sufficiently large \(N\) such that \(U_i = \mathbb{R}\) for all \(i > N\). Construct the sequence \(y = (x_1, x_2, \dots, x_N, 0, 0, \dots)\). Clearly \(y \in R_{ez}^\omega\) and \(y \in U\). This means every point is a limit point, so the closure is the entire space: \(\overline{R_{ez}^\omega} = \mathbb{R}^\omega\).
2. Under the Box Topology \(\mathcal{T}_b\)
-
Interior \((R_{ez}^\omega)^\circ\): Similarly, any non-empty basic open element in the box topology still allows components to take values freely within an interval. As long as \(U_i\) is not a singleton \(\{0\}\), we can always construct a sequence with infinitely many non-zero terms that falls into this neighborhood. Thus, the interior remains empty: \((R_{ez}^\omega)^\circ = \emptyset\).
-
Closure \(\overline{R_{ez}^\omega}\): Basis elements of the box topology take the form \(\prod U_i\), where every component can be restricted. Let \(x\) be a sequence with infinitely many non-zero components (i.e., \(x \notin R_{ez}^\omega\)). For each non-zero component \(x_i\), we can choose an open interval \(U_i = (x_i - \frac{|x_i|}{2}, x_i + \frac{|x_i|}{2})\) that does not contain \(0\). For components that are \(0\), we take any neighborhood. The box open set \(\prod U_i\) constructed this way has an empty intersection with \(R_{ez}^\omega\). Therefore, points not in \(R_{ez}^\omega\) are not limit points, making it a closed set: \(\overline{R_{ez}^\omega} = R_{ez}^\omega\).
Problem 3
(i) Is the topological space \((\mathbb{R}^\omega, \mathcal{T}_p)\) a metrizable space? (ii) Is it connected?
(Hint: For metrizability, consider the function \(d^\omega(x, y) = \sup \{ \frac{d^*(x_i, y_i)}{i+1} : i \in \mathbb{N} \}\), where \(d\) is the standard metric on \(\mathbb{R}\).)
Solution
(i) Metrizability
Conclusion: It is metrizable.
Reason: Using the metric in the hint:
Since the denominator \(i+1\) approaches infinity as the dimension increases, this ensures the metric is very loose on far-end coordinates and very strict on near-end coordinates. This decay characteristic perfectly corresponds to the "only constraints on finite coordinates" structure of the product topology. Through rigorous verification of ball inclusions, it can be proven that the topology induced by this metric is exactly equal to the product topology \(\mathcal{T}_p\).
(ii) Connectivity
Conclusion: It is connected.
Reason: The real line \(\mathbb{R}\) is a connected space under the standard topology. According to a famous theorem in topology: Any Cartesian product of connected spaces (under the product topology) remains connected. Since \(\mathbb{R}^\omega\) is a product of countably many connected spaces, it is connected under the product topology.
Problem 4
Consider the topological space \((\mathbb{R}^\omega, \mathcal{T}_b)\). Let \(U \subseteq \mathbb{R}^\omega\) be the subset consisting of all bounded sequences, i.e.:
Prove that \(U\) and \(\mathbb{R}^\omega - U\) form a non-trivial separation of the space, and deduce whether \((\mathbb{R}^\omega, \mathcal{T}_b)\) is connected.
Solution
1. Prove \(U\) is an open set
For any bounded sequence \(x = (x_i) \in U\), every term satisfies \(|x_i| \le C_x\). Consider the box open set:
For any element \(y = (y_i)\) in \(V\):
Thus \(y\) is also a bounded sequence, meaning \(V \subseteq U\). Since every point has an open neighborhood entirely contained in \(U\), \(U\) is an open set.
2. Prove \(\mathbb{R}^\omega - U\) is an open set
For any unbounded sequence \(z = (z_i) \in \mathbb{R}^\omega - U\), consider the same form of box open set:
If there exists some \(y \in W\) that is bounded (i.e., \(|y_i| \le M\)), then based on the distance relationship:
This would mean \(z\) is also a bounded sequence, contradicting the premise. Therefore, all sequences in \(W\) must be unbounded, i.e., \(W \subseteq \mathbb{R}^\omega - U\). Thus \(\mathbb{R}^\omega - U\) is also an open set.
3. Conclusion
Since both \(U\) and \(\mathbb{R}^\omega - U\) are non-empty open sets, they are disjoint, and their union is the entire space. This constitutes a non-trivial separation of the space. Therefore, the box topology space \((\mathbb{R}^\omega, \mathcal{T}_b)\) is disconnected.
Problem 5
Consider the following subset in \(\mathbb{R}^2\) endowed with the standard subspace topology:
Question: Is the space \(X\) connected? Please provide a proof.
Solution
Conclusion: \(X\) is disconnected.
Proof:
We split \(X\) into two natural geometric parts:
Clearly \(A\) and \(B\) are non-empty sets, \(A \cap B = \emptyset\), and \(X = A \cup B\). To prove \(X\) is disconnected, we only need to show that \(A\) and \(B\) are both closed sets in the subspace \(X\) (thereby forming a non-trivial separation).
-
Proving \(A\) is closed: \(A\) is a complete line, which is a closed set in the standard topology of the entire space \(\mathbb{R}^2\). By the properties of subspace topology, \(A = A \cap X\), so \(A\) must be closed in the subspace \(X\).
-
Proving \(B\) is closed: We need to examine the limit points of \(B\) in the entire space \(\mathbb{R}^2\). Suppose there is a sequence \((x_n, 1/x_n) \in B\) that converges to some point \((x_0, y_0)\) in \(\mathbb{R}^2\).
-
If \(x_0 > 0\), since \(f(x) = 1/x\) is continuous, then \(y_0 = 1/x_0\), and the limit point \((x_0, y_0)\) is still in \(B\).
- If \(x_0 = 0\), as \(x_n \to 0^+\), \(1/x_n \to +\infty\). Since the \(y\)-coordinate diverges, the sequence does not converge in \(\mathbb{R}^2\), and no limit point exists.
- If \(x_0 < 0\), since \(x_n > 0\) in the sequence, it cannot converge to a negative number.
This shows that \(B\) contains all its limit points in \(\mathbb{R}^2\), so \(B\) is itself a closed set in \(\mathbb{R}^2\). Consequently, in the subspace \(X\), \(B = B \cap X\) is also closed.
Since \(X = A \cup B\), and \(A, B\) are non-empty, disjoint closed sets, they constitute a non-trivial separation of \(X\). Therefore, the topological space \(X\) is disconnected.
Problem 6
Prove that the following subspace of \(\mathbb{R}^2\) (a variation of the Topologist's Sine Curve) is not path-connected:
Solution
We use proof by contradiction. Suppose \(X\) is path-connected.
1. Construct a path:
Since \(X\) is path-connected, there must exist a continuous mapping \(\gamma: [0, 1] \to X\) between the origin \((0,0)\) and a point on the curve \((1/\pi, 0)\) such that:
2. Consider component functions:
Let \(\gamma(t) = (x(t), y(t))\). Since \(\gamma\) is a continuous mapping, its coordinate component functions \(x(t)\) and \(y(t)\) must be real-valued continuous functions on the closed interval \([0, 1]\).
Consider the set \(T = \{ t \in [0, 1] \mid x(t) = 0 \}\). Because the singleton \(\{0\}\) is closed and \(x(t)\) is continuous, \(T\) is a closed subset of \([0, 1]\). Let \(t_0 = \sup T\). Since we know \(x(1) = 1/\pi > 0\), it must be that \(t_0 < 1\).
By the properties of the supremum, to the right of \(t_0\) (i.e., for \(t \in (t_0, 1]\)), we have \(x(t) > 0\).
3. Derive a contradiction:
As \(t\) approaches \(t_0\) from the right (\(t \to t_0^+\)), we have \(x(t) \to 0^+\). Since \(\gamma\) lies in \(X\), when \(x(t) > 0\), it must satisfy:
Since \(y(t)\) is also a continuous function, its right-hand limit at \(t_0\) must exist and equal \(y(t_0)\), i.e., \(y(t_0) = 0\). However, as \(x(t) \to 0^+\), the variable \(1/x(t) \to +\infty\). The sine function \(\sin(1/x(t))\) will oscillate infinitely many times between \([-1, 1]\), and no limit exists.
This creates an irreconcilable contradiction with the requirement that \(y(t)\) must be continuous at \(t_0\). Therefore, no such continuous path \(\gamma\) exists. Thus \(X\) is not path-connected.
Problem 7
Prove that the three spaces \((0, 1)\), \((0, 1]\), and \([0, 1]\) are not homeomorphic to each other.
Solution
We distinguish these three subspaces using topological invariants (compactness and cut-point properties):
1. Distinguishing \([0, 1]\) from the others
Under the standard topology, the closed interval \([0, 1]\) is a bounded closed set in the real line. According to the Heine-Borel Theorem, \([0, 1]\) is a compact space. In contrast, the open interval \((0, 1)\) and the semi-open interval \((0, 1]\) are not closed sets, so they are not compact. Since homeomorphism must preserve compactness, \([0, 1]\) is not homeomorphic to either \((0, 1)\) or \((0, 1]\).
2. Distinguishing \((0, 1)\) from \((0, 1]\)
We use the number of cut points for topological differentiation.
- For \((0, 1]\), there exists a special point (the endpoint \(1\)). If we remove this point, the remaining space \((0, 1] \setminus \{1\} = (0, 1)\) is still a connected interval. Thus, \((0, 1]\) has a non-cut point.
- For \((0, 1)\), every point is a cut point. Regardless of which point \(c \in (0, 1)\) is removed, the space breaks into two disjoint non-empty open sets \((0, c) \cup (c, 1)\), making it disconnected.
A homeomorphism must preserve the topological structure of cut points and non-cut points. Since their cut-point properties are fundamentally different, \((0, 1)\) and \((0, 1]\) are not homeomorphic.
In conclusion, these three spaces are pairwise non-homeomorphic.
Problem 8
(i) Prove that every continuous function \(f: [0, 1] \to [0, 1]\) has at least one fixed point \(x \in [0, 1]\), i.e., \(f(x) = x\). (ii) If \([0, 1]\) is replaced by \((0, 1)\) or \((0, 1]\), does the above conclusion still hold? Please provide reasons.
Solution
(i) Proving the existence of a fixed point
We construct an auxiliary function \(g(x) = f(x) - x\). Since \(f(x)\) is continuous and the identity map \(x\) is also continuous, \(g(x)\) is a continuous function on the closed interval \([0, 1]\).
Examine the values at the endpoints:
- If \(f(0) = 0\) or \(f(1) = 1\), then a fixed point is already found.
- Otherwise, it must be that \(f(0) > 0\) (since the range of \(f\) is in \([0, 1]\)) and \(f(1) < 1\).
- Substituting into the auxiliary function gives \(g(0) = f(0) - 0 > 0\) and \(g(1) = f(1) - 1 < 0\).
According to the Intermediate Value Theorem for continuous real functions, since the function values have opposite signs at the endpoints, there must exist a point \(x_0\) in the interior \((0, 1)\) such that:
Which means \(f(x_0) - x_0 = 0 \implies f(x_0) = x_0\). Thus \(f\) must have at least one fixed point on \([0, 1]\).
(ii) Discussion on other intervals
Conclusion: For \((0, 1)\) and \((0, 1]\), the conclusion does not hold.
-
Counter-example for \((0, 1)\): Construct the continuous function \(f(x) = x^2\). It maps \((0, 1)\) to \((0, 1)\). Solving \(x^2 = x\) yields \(x=0\) or \(x=1\), but neither point is in the open interval \((0, 1)\). Therefore, this function has no fixed point. (Or take \(f(x) = (x+1)/2\), where the solution \(x=1\) is not in the interval).
-
Counter-example for \((0, 1]\): Construct the continuous function \(f(x) = x/2\). It maps \((0, 1]\) to \((0, 1/2] \subseteq (0, 1]\). Solving \(x/2 = x\), the only solution is \(x=0\), which is not inside the interval \((0, 1]\). Thus, there is no fixed point.
Problem 9
Let \((X, \mathcal{T})\) be a topological space. An open neighborhood basis of a point \(x \in X\) is a collection of open sets \(\{B_i\}_{i \in I}\) satisfying:
- For all \(i\), \(x \in B_i\);
- For any open set \(U\) containing \(x\), there exists some \(B_i\) such that \(B_i \subseteq U\).
If every point in a topological space \(X\) has an open neighborhood basis consisting of path-connected sets, we say \(X\) is locally path-connected.
Prove: If \(X\) is connected and locally path-connected, then \(X\) must be path-connected. (Hint: Fix \(x \in X\), and consider the set \(S_x\) of all points in \(X\) that can be connected to \(x\) by a path. Prove that \(S_x\) is both open and closed.)
Solution
1. Construct the path-component set \(S_x\)
Fix any point \(x \in X\). We define the set \(S_x\) as the set of points that can be connected to \(x\) by a path:
A constant path (identically equal to \(x\)) clearly connects \(x\) to itself, so \(x \in S_x\), making \(S_x\) a non-empty set.
2. Prove \(S_x\) is an open set
Take any \(y \in S_x\). Since \(X\) is locally path-connected, for point \(y\) and the space \(X\), there must exist a path-connected open set \(B_y\) from \(y\)'s neighborhood basis such that \(y \in B_y \subseteq X\).
Examine any point \(z\) in \(B_y\):
- Since \(y \in S_x\), there is a path from \(x\) to \(y\).
- Since \(B_y\) is path-connected, there is a path from \(y\) to \(z\).
Concatenating these two paths gives a path from \(x\) to \(z\), implying \(z \in S_x\). Since \(z \in S_x\) for every \(z \in B_y\), then \(B_y \subseteq S_x\). This means every point in \(S_x\) has an open neighborhood entirely contained within \(S_x\), so \(S_x\) is an open set.
3. Prove \(S_x\) is a closed set
We need to show that its complement \(X \setminus S_x\) is open.
Take any \(w \in X \setminus S_x\). Similarly, by local path-connectivity, \(w\) must have a path-connected open neighborhood \(B_w\). If this neighborhood \(B_w\) intersects \(S_x\), suppose there is a point \(v \in B_w \cap S_x\):
- \(v \in S_x\) implies there is a path from \(x\) to \(v\).
- \(B_w\) being path-connected implies there is a path from \(v\) to \(w\).
Concatenating the paths would imply \(w \in S_x\), which contradicts \(w \in X \setminus S_x\)! Thus, \(B_w \cap S_x = \emptyset\), meaning \(B_w \subseteq X \setminus S_x\). Since every point in the complement has an open neighborhood in the complement, \(X \setminus S_x\) is open, so \(S_x\) is closed.
4. Conclusion
In summary, \(S_x\) is a non-empty, both open and closed set in the topological space \(X\). Since the space \(X\) is connected, the only non-empty set that is both open and closed must be \(X\) itself. Therefore:
This indicates that every point in the space can be connected to \(x\) by a path, so \(X\) must be path-connected.
Problem 10
Consider the finite-complement topology on \(\mathbb{R}\): \(\mathcal{T} = \{ \emptyset \} \cup \{ \mathbb{R} \setminus A : A \text{ is a finite set} \}\).
Prove: Every subset \(X \subseteq \mathbb{R}\) (endowed with the subspace topology) is compact.
Solution
Proof:
Let \(\{U_i\}_{i \in I}\) be an arbitrary open cover of the subspace \(X\).
-
If \(X\) is an empty set, it is trivially compact.
-
If \(X\) is non-empty, we can pick any non-empty open set from the cover, say \(U_{i_0}\).
-
According to the definitions of subspace topology and finite-complement topology, \(U_{i_0}\) must take the form \(U_{i_0} = X \cap (\mathbb{R} \setminus A)\), where \(A\) is some finite set in \(\mathbb{R}\).
-
Now, we examine the remaining part of \(X\) not yet covered by \(U_{i_0}\): \(X \setminus U_{i_0} = X \setminus (X \cap (\mathbb{R} \setminus A)) = X \cap A\). Since \(A\) is a finite set, the intersection \(X \cap A\) must also be a finite set. This means there are only finitely many points in \(X\) not covered by \(U_{i_0}\); let's call them \(\{x_1, x_2, \dots, x_k\}\).
-
Since \(\{U_i\}_{i \in I}\) covers \(X\), for each remaining point \(x_j\) (\(j = 1, \dots, k\)), there must exist some open set \(U_{i_j}\) in the cover such that \(x_j \in U_{i_j}\).
-
Thus, we only need to take the initially chosen \(U_{i_0}\) plus these \(k\) extra open sets to form a collection: \(\{U_{i_0}, U_{i_1}, \dots, U_{i_k}\}\).
Clearly, this is a finite subcollection of open sets that completely covers \(X\).
Since every open cover of \(X\) has a finite subcover, by the definition of compactness, the subspace \(X\) must be compact.