Exercise 4 (Homework 07 & 08 & 09)
This section contains the seventh homework assignment of the Topology course and its related proofs. It covers the separation of Hausdorff spaces and compact sets, the cardinality of connected metric spaces, a special case of the Baire category theorem, topological properties of the Cantor set, and the minimality of compact topological spaces.
Problem 1
Let \((X, \mathcal{T})\) be a Hausdorff space. Let \(A, B \subseteq X\) be two disjoint compact subsets.
Prove that there exist disjoint open sets \(U, V\) in \(X\) such that \(A \subseteq U\) and \(B \subseteq V\).
Solution
Proof:
First fix an arbitrary point \(a \in A\).
For every point \(b \in B\), since \(A \cap B = \emptyset\), we necessarily have \(a \ne b\).
Since \(X\) is Hausdorff, for distinct points \(a\) and \(b\), there exist disjoint open sets \(U_{a,b}\) and \(V_{a,b}\) in \(X\) such that \(a \in U_{a,b}\) and \(b \in V_{a,b}\).
As \(b\) ranges over the whole set \(B\), the family of open sets \(\{V_{a,b}\}_{b \in B}\) forms an open cover of the compact set \(B\).
By compactness, there is a finite subcover; that is, there exist finitely many points \(b_1, b_2, \dots, b_n \in B\) such that:
For these finitely many points, define:
Clearly \(U_a\) is a finite intersection of open sets, hence open; \(V_a\) is a union of open sets, hence also open. Moreover, since \(U_{a,b_i} \cap V_{a,b_i} = \emptyset\) for all \(i\), we deduce \(U_a \cap V_a = \emptyset\). At this point, \(a \in U_a\) and \(B \subseteq V_a\).
Now let \(a\) range over the whole set \(A\). The family of open sets \(\{U_a\}_{a \in A}\) forms an open cover of the compact set \(A\).
Similarly, there exist finitely many points \(a_1, a_2, \dots, a_m \in A\) such that the finite subcover holds:
Finally, define:
Clearly \(U\) and \(V\) are open sets and satisfy \(A \subseteq U\) and \(B \subseteq V\).
Check that they are disjoint:
Therefore, we have found disjoint open sets \(U\) and \(V\) satisfying the required conditions.
Problem 2
Prove that a connected metric space \((X, d)\) containing more than one point must be uncountable.
(Hint: Pick \(a \in X\) and consider the function \(d_a: X \to \mathbb{R}\), defined by \(d_a(x) := d(a, x)\). Think about the connectedness of countable subsets of \(\mathbb{R}\).)
Solution
Proof:
Since the space \(X\) contains more than one point, there exist two points \(a, b \in X\) such that \(a \ne b\). By positivity of the metric, we have:
Fix the point \(a\) and define the distance function \(d_a: X \to \mathbb{R}\) by:
By the triangle inequality, for any \(x, y \in X\), we have \(|d_a(x) - d_a(y)| = |d(a, x) - d(a, y)| \le d(x, y)\). Therefore \(d_a\) is (uniformly) continuous on \(X\).
We know the following important property of topological spaces: the image of a connected space under a continuous map is connected. Since \(X\) is connected, the image \(d_a(X)\) must be a connected subset of the real line \(\mathbb{R}\).
Examine the elements of \(d_a(X)\). Since \(a \in X\) and \(b \in X\), the image must contain:
In \(\mathbb{R}\), a connected subset containing both \(0\) and \(d(a,b)\) must be an interval containing the entire closed interval \([0, d(a,b)]\).
Any real interval of positive length is uncountable. Hence the image \(d_a(X)\) is uncountable.
Since the image of a function is uncountable, its domain \(X\) must also be uncountable.
Problem 3
(A special case of the Baire category theorem) Let \((X, \mathcal{T})\) be a compact Hausdorff space. Let \(\{A_i : i \in \mathbb{N}\}\) be a family of closed subsets of \(X\), and suppose that for each \(i\), its interior is empty, i.e. \((A_i)^\circ = \emptyset\).
Prove that the interior of the union is empty, i.e. \(\left( \bigcup_{i=1}^\infty A_i \right)^\circ = \emptyset\).
Solution
Proof:
We argue by contradiction. Suppose the interior of the union is nonempty. Then there exists a nonempty open set \(V_0\) such that:
Consider the set \(A_1\). Since \(A_1\) is closed and has empty interior (that is, it contains no nonempty open set), the open set \(V_0\) can never be completely contained in \(A_1\). Therefore, the set \(V_0 \setminus A_1\) is a nonempty open set.
A compact Hausdorff space is necessarily a regular space. In a regular space, every nonempty open set contains the closure of some nonempty open set. Hence there exists a nonempty open set \(V_1\) satisfying:
Next consider \(A_2\). Similarly, since \(A_2\) has empty interior, \(V_1\) cannot be completely contained in \(A_2\), so \(V_1 \setminus A_2\) is also nonempty open. By regularity, we can find a nonempty open set \(V_2\) satisfying:
Repeating this process indefinitely, we construct a sequence of open sets \(\{V_n\}\) such that for every \(k \ge 1\):
Thus we obtain a nested decreasing sequence of nonempty closed sets:
Since \(X\) is compact and this family of closed sets has the finite intersection property, the intersection of the entire decreasing sequence must be nonempty:
Let \(x \in \bigcap_{k=1}^\infty \overline{V_k}\). Since all the \(\overline{V_k}\) are contained in \(V_0\), we have \(x \in V_0\). By the initial assumption \(V_0 \subseteq \bigcup A_i\), there exists some integer \(m\) such that \(x \in A_m\).
However, by the construction of the sequence, \(x \in \overline{V_m}\). And by the recursive construction:
This implies \(x \notin A_m\), which contradicts the conclusion in step 8 that \(x \in A_m\).
This contradiction shows that the initial assumption was false. Therefore the union cannot contain any nonempty open set, i.e. \(\left( \bigcup_{i=1}^\infty A_i \right)^\circ = \emptyset\).
Problem 4
(The Cantor set) Let \(A_0 = [0, 1]\), and for each \(n \in \mathbb{N}\), define recursively:
Let \(C := \bigcap_{n \ge 0} A_n\), endowed with the standard subspace topology inherited from \(\mathbb{R}\).
Prove the following: (i) \(C\) is totally disconnected (i.e. every connected component of \(C\) is a singleton). (ii) \(C\) is compact and has no isolated points, and deduce that \(C\) is uncountable.
Solution
(i) Prove that \(C\) is totally disconnected
Suppose, to the contrary, that \(C\) has a connected component \(S\) containing more than one point.
Take two points \(x, y \in S\) with \(x < y\). Since \(x, y \in S \subseteq C\), we must have \(x, y \in A_n\) for all \(n \ge 0\).
Choose \(n\) large enough so that the length of the removed intervals satisfies:
At the \(n\)-th stage of the construction, \(A_n\) consists of a union of disjoint closed intervals of length \(\frac{1}{3^n}\).
Since \(y - x > 1/3^n\), the points \(x\) and \(y\) cannot lie in the same closed interval of \(A_n\).
Therefore, between \(x\) and \(y\) there must be at least one open interval that was removed during the construction (e.g. the middle third). Let \(z\) be a point in one such removed interval. Then \(x < z < y\) and \(z \notin C\).
But a connected subset \(S\) of the real line has the property that if \(x, y \in S\), then the entire interval \([x, y]\) must be contained in \(S\). This contradicts the fact that \(z \notin C\). Therefore every connected component of \(C\) is a singleton.
(ii) Compactness, absence of isolated points, and uncountability
Compactness: Each \(A_n\) is a finite union of closed intervals, hence closed in \([0, 1]\). The intersection \(C = \bigcap A_n\) of arbitrarily many closed sets is still closed in \([0, 1]\). Since \([0, 1]\) is a closed and bounded subset of \(\mathbb{R}\) (hence compact), its closed subset \(C\) is necessarily compact.
No isolated points: Take any \(x \in C\) and any relatively open neighbourhood \(V\) of \(x\) in \(C\). There exists an open set \(U\) in \(\mathbb{R}\) such that \(V = U \cap C\). Since \(U\) is open, there exists \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon) \subseteq U\).
For every \(n\), \(x \in A_n\), so \(x\) lies in one of the closed intervals \(I_n\) of \(A_n\); this interval has length \(1/3^n\).
By construction, the two endpoints of \(I_n\) are never removed at later stages, so they both belong to \(C\).
Choose \(n\) large enough so that \(1/3^n < \epsilon\). Then the whole interval \(I_n\) is contained in \((x - \epsilon, x + \epsilon)\).
The two endpoints of \(I_n\) belong to \(V\). Even if \(x\) happens to be one of these endpoints, the other endpoint is distinct from \(x\) and belongs to \(V\). This shows that every neighbourhood of \(x\) contains a point of \(C\) different from \(x\). Hence \(x\) is not an isolated point (i.e. \(C\) is perfect).
Uncountability: We have shown that \(C\) is compact (hence a complete metric space) with no isolated points. By the Baire category theorem, since a countable union of nowhere dense singleton sets cannot constitute a complete space, \(C\) must be uncountable.
Problem 5
Fix a compact topological space \((X, \mathcal{T})\). Prove that the following two statements are equivalent:
(i) With respect to \(\mathcal{T}\), every compact subset of \(X\) is closed in \(X\); (ii) For any topology \(\mathcal{T}'\) strictly finer than \(\mathcal{T}\), the space \((X, \mathcal{T}')\) is non-compact.
Also, give an example of a space \((X, \mathcal{T})\) satisfying the above (equivalent) conditions.
Solution
Proof of (i) \(\implies\) (ii)
Let \(\mathcal{T}'\) be a topology strictly finer than \(\mathcal{T}\) (i.e. \(\mathcal{T}' \supsetneq \mathcal{T}\)).
Because the topology is finer, there exists an open set \(U \in \mathcal{T}' \setminus \mathcal{T}\).
Let \(K = X \setminus U\). Clearly \(K\) is closed in \(\mathcal{T}'\), but not closed in \(\mathcal{T}\).
We argue by contradiction. Suppose the space \((X, \mathcal{T}')\) is still compact.
Since \(K\) is a closed subset of the compact space \((X, \mathcal{T}')\), by standard topological properties \(K\) is compact in \((X, \mathcal{T}')\).
Consider the identity map \(id: (X, \mathcal{T}') \to (X, \mathcal{T})\). Since \(\mathcal{T}'\) contains \(\mathcal{T}\), this map is continuous.
The continuous image of a compact set is compact. Therefore \(K\) must also be compact in \((X, \mathcal{T})\).
By assumption (i), every compact subset of \((X, \mathcal{T})\) must be closed. Thus \(K\) must be closed in \(\mathcal{T}\).
This contradicts the conclusion in step 3 that \(K\) is not closed in \(\mathcal{T}\). Therefore \((X, \mathcal{T}')\) cannot be compact.
Proof of (ii) \(\implies\) (i)
Again we argue by contradiction. Suppose condition (i) is false; that is, there exists a compact subset \(K\) of \(X\) which is not closed in \(\mathcal{T}\).
Since \(K\) is not closed, its complement \(X \setminus K\) is not open in \(\mathcal{T}\).
We construct a new topology \(\mathcal{T}'\) by adding \(X \setminus K\) to the family of open sets. Let its subbasis be \(\mathcal{T} \cup \{X \setminus K\}\). Since \(X \setminus K \notin \mathcal{T}\), the new topology \(\mathcal{T}'\) is strictly finer than \(\mathcal{T}\).
We now show that under this finer topology, \((X, \mathcal{T}')\) is still compact.
Let \(\{U_\alpha\} \cup \{V_\beta \setminus K\}\) be an arbitrary open cover of \(X\) in the new topology \(\mathcal{T}'\), where \(U_\alpha, V_\beta \in \mathcal{T}\).
Since sets of the form \(V_\beta \setminus K\) cannot cover any point of \(K\), the subset \(K\) must be entirely covered by the original open sets \(\{U_\alpha\}\).
Since \(K\) is compact in \(\mathcal{T}\), there exists a finite subcover \(\{U_{\alpha_1}, \dots, U_{\alpha_n}\}\) covering \(K\).
Let \(W = \bigcup_{i=1}^n U_{\alpha_i}\). Then \(W\) is a \(\mathcal{T}\)-open set containing \(K\).
Consider the remaining uncovered part \(X \setminus W\). It is \(\mathcal{T}\)-closed and contained in the compact space \(X\). Hence \(X \setminus W\) is \(\mathcal{T}\)-compact.
Also note that \(X \setminus W \subseteq X \setminus K\), so it is covered by the remaining sets in the cover. Using the finite subcover property, only finitely many of these sets are needed to cover \(X \setminus W\).
Hence we have extracted a finite subcover of the whole space \(X\), proving that \((X, \mathcal{T}')\) is compact.
But this contradicts condition (ii), which says every strictly finer topology must be non-compact. Therefore the assumption is false and condition (i) must hold.
Example of a space satisfying the conditions
Any compact Hausdorff space satisfies these conditions.
A concrete example: the closed interval \([0, 1]\) with the standard topology.
Reason: in a Hausdorff space, every compact subset is necessarily closed, so it satisfies condition (i) and therefore equivalently condition (ii).
Problem 6
Determine whether the following two spaces are locally compact:
(i) The rational numbers \(\mathbb{Q}\) with the standard topology; (ii) The real line \((\mathbb{R}, \mathcal{T})\) with the topology generated by the basis \(\mathcal{B} = \{(a, b] : a, b \in \mathbb{R}, a < b\}\).
Justify your answers.
Solution
(i) \(\mathbb{Q}\) with the standard topology is not locally compact
Proof: We argue by contradiction. Suppose \(\mathbb{Q}\) is locally compact. Then for any point \(x \in \mathbb{Q}\), there must exist a compact neighbourhood \(K\) of \(x\) in \(\mathbb{Q}\).
Since \(K\) is a neighbourhood of \(x\), there exists \(\epsilon > 0\) such that the open set \(U = (x - \epsilon, x + \epsilon) \cap \mathbb{Q}\) satisfies:
Since \(\mathbb{Q}\) is a metric space, its compact subset \(K\) must be sequentially compact.
Now choose an irrational number \(y\) in the interval \((x - \epsilon, x + \epsilon)\) (such irrationals exist because they are dense in \(\mathbb{R}\)). By the density of the rationals, we can construct a sequence of rationals \(\{x_n\} \subset U \subseteq K\) such that \(x_n \to y\) in the usual metric on \(\mathbb{R}\).
Since \(K\) is sequentially compact, the sequence \(\{x_n\}\) must have a convergent subsequence within \(K\) (hence within \(\mathbb{Q}\)). However, in \(\mathbb{R}\) the sequence converges uniquely to the irrational number \(y \notin \mathbb{Q}\), so in \(\mathbb{Q}\) the sequence has no limit point. This is a fundamental contradiction to the sequential compactness of \(K\). Hence \(\mathbb{Q}\) is not locally compact.
(ii) \((\mathbb{R}, \mathcal{T})\) with the upper limit topology is not locally compact
Proof: The basic open sets of this topology are left-open right-closed intervals \((a, b]\). We again argue by contradiction. Suppose this space is locally compact. Then any point \(x \in \mathbb{R}\) must have a compact neighbourhood \(K\) containing \(x\).
By definition of neighbourhood, there exists a basis element \((a, x]\) such that:
Notice that in the upper limit topology, the set \((a, x]\) is both open and closed (its complement \((-\infty, a] \cup (x, \infty)\) can be written as a union of basis elements and is open). A closed subset of a compact set \(K\) must also be compact, so \((a, x]\) must be compact, and thus must be limit point compact.
We construct a sequence in \((a, x]\) that decreases strictly to \(a\), for example:
Clearly \(x_n \in (a, x]\) for all \(n\). For any point \(y \in (a, x]\), we may choose a neighbourhood of the form \((c, y]\) in the upper limit topology. Since the sequence \(x_n\) decreases strictly to \(a < y\), for sufficiently large \(n\) we have \(x_n < c < y\), meaning that this neighbourhood \((c, y]\) contains only finitely many points of the sequence.
Therefore this sequence has no limit point in \((a, x]\). This contradicts the fact that \((a, x]\), being compact, must be limit point compact. Hence this space is not locally compact.
Problem 7
Prove that the product space \(\left(\prod_{i=1}^{\infty} A_i, \mathcal{T}_p\right)\) is compact in the product topology, where each \(A_i\) is a nonempty finite set with the discrete topology.
(Hint: Review the conclusion from Homework 5; consider sequential compactness and metrizability.)
Solution
Proof:
First, since each \(A_i\) is finite and discrete, it is a metrizable space. The product of countably many metrizable spaces is metrizable in the product topology. For metric spaces, compactness is equivalent to sequential compactness. Therefore, it suffices to prove that the product space is sequentially compact.
Let \(\{x^{(n)}\}_{n=1}^{\infty}\) be an arbitrary sequence in the product space \(\prod_{i=1}^{\infty} A_i\). We will use the Cantor diagonal argument to extract a convergent subsequence.
Consider the sequence of first coordinates \(\{x_1^{(n)}\}_{n=1}^{\infty}\). Since \(A_1\) is finite and the sequence has infinitely many terms, by the pigeonhole principle there is some element of \(A_1\) that is taken infinitely many times. Hence we can extract a subsequence \(\{x^{(n, 1)}\}\) from the original sequence such that its first coordinate is constant.
Next, consider the second coordinates of the subsequence \(\{x^{(n, 1)}\}\). Since \(A_2\) is finite, we can extract from \(\{x^{(n, 1)}\}\) a further subsequence \(\{x^{(n, 2)}\}\) whose second coordinate is also constant (now both the first and second coordinates are fixed).
Continuing inductively, suppose we have extracted the \(k\)-th subsequence \(\{x^{(n, k)}\}\). Since \(A_{k+1}\) is finite, we can always extract the \((k+1)\)-th subsequence \(\{x^{(n, k+1)}\}\) whose first \(k+1\) coordinates are all constant.
Now construct the diagonal sequence:
Consider this diagonal sequence \(y^{(m)}\). For any fixed coordinate position \(k\), when \(m \ge k\), \(y^{(m)}\) belongs to the \(k\)-th subsequence (or a later subsequence), so its \(k\)-th coordinate is a fixed constant.
In the product topology, convergence of a sequence is equivalent to coordinate-wise convergence. Hence for every fixed \(k\), the \(k\)-th coordinate of the diagonal sequence \(y^{(m)}\) is eventually constant, so it converges in the \(k\)-th coordinate.
Therefore the diagonal sequence \(y^{(m)}\) converges in the product topology to some point of the space. This shows that every sequence has a convergent subsequence, so the product space is sequentially compact, and therefore compact.
Problem 8
(\(p\)-adic integers) Fix a prime number \(p\). For each \(i \in \mathbb{N}_{\ge 1}\), we have the quotient ring \(\mathbb{Z}/p^i\mathbb{Z}\) induced by the equivalence relation. Consider the following set:
where \(\phi_i: \mathbb{Z}/p^i\mathbb{Z} \to \mathbb{Z}/p^{i-1}\mathbb{Z}\) is the natural surjection. Give each \(\mathbb{Z}/p^i\mathbb{Z}\) the discrete topology, and give the product space the product topology.
(i) Prove that \(\mathbb{Z}_p\) is closed in the product space \(\prod_{i=1}^{\infty} (\mathbb{Z}/p^i\mathbb{Z})\), and is compact. (ii) For each \(i \ge 1\), let \(f_i: \mathbb{Z} \to \mathbb{Z}/p^i\mathbb{Z}\) be the standard quotient map. Consider the embedding \(f: \mathbb{Z} \hookrightarrow \mathbb{Z}_p\), defined by \(f(a) = (f_i(a))_{i \ge 1}\). Prove that \(f(\mathbb{Z})\) is dense in \(\mathbb{Z}_p\).
Solution
(i) Prove that \(\mathbb{Z}_p\) is closed and compact
For any fixed \(i \ge 1\), define the map \(\psi_i: \prod_{k=1}^{\infty} (\mathbb{Z}/p^k\mathbb{Z}) \to \mathbb{Z}/p^i\mathbb{Z}\) by:
This map \(\psi_i\) depends only on the \(i\)-th and \((i+1)\)-th coordinates of the product space. Since projection maps are continuous and algebraic operations on finitely many coordinates are continuous in the discrete topology, \(\psi_i\) must be continuous.
Let \(E_i\) be the set defined by the \(i\)-th compatibility condition:
Since the singleton \(\{0\}\) is closed in the discrete topology and \(\psi_i\) is continuous, the preimage \(E_i\) is closed in the product topology.
The space \(\mathbb{Z}_p\) of \(p\)-adic integers is the intersection of all compatibility sets:
An arbitrary intersection of closed sets is closed, hence \(\mathbb{Z}_p\) is closed.
On the other hand, each coordinate space \(\mathbb{Z}/p^i\mathbb{Z}\) is finite and hence compact in the discrete topology. By the Tychonoff theorem, the product space \(\prod_{i=1}^{\infty} (\mathbb{Z}/p^i\mathbb{Z})\) is compact.
A closed subset of a compact space is compact. Since \(\mathbb{Z}_p\) is closed in the compact product space, \(\mathbb{Z}_p\) itself is compact.
(ii) Prove that \(f(\mathbb{Z})\) is dense
To prove density, it suffices to show that for every nonempty basic open set \(U\) in the subspace topology of \(\mathbb{Z}_p\), we have \(U \cap f(\mathbb{Z}) \ne \emptyset\).
Under the product topology, a basic open set \(U\) in \(\mathbb{Z}_p\) is determined by finitely many coordinate restrictions. Because of the compatibility conditions \(\phi_{i+1}(a_{i+1}) = a_i\) in \(\mathbb{Z}_p\), higher-order coordinates completely determine the lower-order coordinates.
Hence every nonempty basic open set in \(\mathbb{Z}_p\) can be represented by a restriction on one sufficiently large \(N\)-th coordinate: there exists a positive integer \(N\) and a constant \(c \in \mathbb{Z}/p^N\mathbb{Z}\) such that:
Since \(c\) is an equivalence class in the quotient ring \(\mathbb{Z}/p^N\mathbb{Z}\), we can choose a representative \(a \in \mathbb{Z}\) such that:
Consider the image \(f(a) = (f_i(a))_{i \ge 1}\) of the integer \(a\) under the map \(f\). By construction, its \(N\)-th coordinate is exactly \(f_N(a) = c\).
Therefore \(f(a) \in U\). This shows that \(U \cap f(\mathbb{Z})\) is nonempty. Since every nonempty open set contains a point of \(f(\mathbb{Z})\), we have \(\overline{f(\mathbb{Z})} = \mathbb{Z}_p\), i.e. \(f(\mathbb{Z})\) is dense in \(\mathbb{Z}_p\).
Problem 9
Consider the subspace \(Y = \{0\} \cup \{1/n : n \in \mathbb{N}_{\ge 1}\}\) of the real line \(\mathbb{R}\). Is \(Y\) homeomorphic to the one-point compactification \(X = (\mathbb{N}_{\ge 1} \cup \{\infty\}, \mathcal{T}')\) of the natural numbers \(\mathbb{N}_{\ge 1}\) with the discrete topology? Justify your answer.
Solution
Conclusion: Yes, they are homeomorphic.
Proof:
We first construct a bijection \(f: X \to Y\) as follows:
This is clearly a bijection.
We now prove that \(f\) is continuous. We need to check that the preimage under \(f\) of every open set in \(Y\) is open in \(X\).
Let \(V\) be an arbitrary open set in \(Y\). Consider two cases:
-
Case 1: \(0 \notin V\). Then \(V\) consists entirely of isolated points of the form \(1/n\). Its preimage \(f^{-1}(V)\) is a subset of \(\mathbb{N}_{\ge 1}\) and does not contain \(\infty\). Since \(\mathbb{N}_{\ge 1}\) is discrete, in its one-point compactification \(X\), every subset not containing \(\infty\) is open. Hence \(f^{-1}(V)\) is open in \(X\).
-
Case 2: \(0 \in V\). Since \(V\) is open in the subspace \(Y\) of \(\mathbb{R}\), there exists \(\epsilon > 0\) such that \((-\epsilon, \epsilon) \cap Y \subseteq V\). This means that for all \(n > 1/\epsilon\), the points \(1/n\) belong to \(V\). In other words, \(V\) misses only finitely many points of \(Y\) of the form \(1/n\). Consider its preimage \(f^{-1}(V)\): it contains the point \(\infty\), and its complement in \(\mathbb{N}_{\ge 1}\) is finite. By the definition of the one-point compactification, a set containing the point at infinity and having finite complement is exactly an open neighbourhood in the compactification space \(X\). Hence \(f^{-1}(V)\) is again open.
This proves that \(f\) is continuous.
Finally, we use an important theorem in general topology: a continuous bijection from a compact space to a Hausdorff space is necessarily a homeomorphism.
- The space \(X\) is a one-point compactification, which by definition is a compact space.
- The space \(Y\) is a subspace of the metric space \(\mathbb{R}\), hence a Hausdorff space.
Since \(f: X \to Y\) is a continuous bijection satisfying the “compact domain to Hausdorff codomain” condition, \(f\) must be a homeomorphism.
Problem 10
Prove that the product topological space \((\mathbb{R}^\omega, \mathcal{T}_p)\) has no one-point compactification.
(Hint: Is \(\mathbb{R}^\omega\) locally compact?)
Solution
Proof:
According to the fundamental theory of topology, a space \(X\) has a Hausdorff one-point compactification if and only if \(X\) itself is locally compact, Hausdorff, and non-compact. The product space \(\mathbb{R}^\omega\) is clearly Hausdorff (since it is a product of Hausdorff spaces \(\mathbb{R}\)). We will prove that it is not locally compact, so it cannot have a one-point compactification.
We argue by contradiction. Suppose \((\mathbb{R}^\omega, \mathcal{T}_p)\) is locally compact.
Consider the origin \(\mathbf{0} = (0, 0, \dots) \in \mathbb{R}^\omega\). By local compactness, the origin must have a compact neighbourhood \(K\).
By definition of the product topology basis, since \(K\) is a neighbourhood of \(\mathbf{0}\), there exists a basic open set \(U\) such that:
In the product topology, a basic open set \(U\) can restrict only finitely many coordinates. That is, there exists a positive integer \(N\) such that:
where the first \(N\) sets \(U_i\) are open intervals in \(\mathbb{R}\), and from the \((N+1)\)-th coordinate onward, all factors must be the whole space \(\mathbb{R}\).
Now consider the projection onto the \((N+1)\)-th coordinate \(\pi_{N+1}: \mathbb{R}^\omega \to \mathbb{R}\). In the product topology the projection map is always continuous.
Since \(K\) is compact and continuous maps send compact sets to compact sets, the projection \(\pi_{N+1}(K)\) must be a compact subset of \(\mathbb{R}\).
However, from the inclusion \(U \subseteq K\), we obtain the corresponding inclusion of projections:
By the structure of the basic open set \(U\), the projection onto the \((N+1)\)-th coordinate is the whole real line:
Combining steps 8 and 9, we get \(\mathbb{R} \subseteq \pi_{N+1}(K)\). Thus the compact set \(\pi_{N+1}(K)\) must equal the whole space \(\mathbb{R}\).
But we know that the real line \(\mathbb{R}\) is unbounded, and therefore not compact! This is an irreconcilable contradiction.
The contradiction shows that our assumption was false, i.e. \(\mathbb{R}^\omega\) is not locally compact. Lacking local compactness, it cannot admit a one-point compactification.
This section contains the ninth homework assignment of the point-set topology course and its related proofs. It covers separability of product spaces, the equivalence of second countability/Lindelöf/separability in metric spaces, the \(T_0\) separation property of pseudometric-generated topologies, properties of \(T_3\) and \(T_4\) spaces, and the regularity of locally compact Hausdorff spaces.
Problem 11
Prove that if \(X_i\) is separable for every \(i \in \mathbb{N}\), then the product space \(\left(\prod_{i \in \mathbb{N}} X_i, \mathcal{T}_p\right)\) is also separable.
Solution
Proof:
Since each \(X_i\) is separable, for each \(i \in \mathbb{N}\) there exists a countable dense subset \(A_i \subseteq X_i\).
In the product space \(\prod X_i\), fix a “base point” \(x^* = (x_1^*, x_2^*, \dots)\).
For each \(n \in \mathbb{N}\), define the set \(D_n\) to be the set of sequences whose first \(n\) coordinates belong to \(A_i\) and whose remaining coordinates coincide with the base point:
Clearly each \(D_n\) is in bijection with the finite product \(A_1 \times A_2 \times \dots \times A_n\). Since a finite product of countable sets is countable, each \(D_n\) is countable.
Let \(D = \bigcup_{n=1}^\infty D_n\). As a countable union of countable sets, \(D\) must be countable.
We now prove that \(D\) is dense in the product topology. Take an arbitrary nonempty basic open set \(U = \prod_{i=1}^\infty U_i\) in the product space.
By definition of the product topology, only finitely many coordinates have \(U_i\) not equal to the whole space. That is, there exists a positive integer \(N\) such that for all \(i > N\), \(U_i = X_i\).
For \(i \le N\), since \(U_i\) is a nonempty open set in \(X_i\) and \(A_i\) is dense in \(X_i\), there exists a point \(y_i \in U_i \cap A_i\).
Construct a sequence \(y = (y_1, y_2, \dots) \in \prod X_i\), where the first \(N\) coordinates are the chosen points \(y_i\), and all coordinates after \(N\) are the base point coordinates \(x_i^*\).
By definition, since only the first \(N\) coordinates of \(y\) may differ from \(x^*\) and these \(N\) coordinates all belong to \(A_i\), we have \(y \in D_N \subseteq D\).
Also, for \(i \le N\), \(y_i \in U_i\); for \(i > N\), \(y_i = x_i^* \in X_i = U_i\). Hence \(y \in U\).
This shows \(U \cap D \ne \emptyset\). Since every basic open set contains a point of \(D\), \(D\) is dense. Having found a countable dense subset, the product space is separable.
Problem 12
Let \((X, d)\) be a metric space. Prove that the following three properties are equivalent:
(i) \(X\) is second countable; (ii) \(X\) is a Lindelöf space; (iii) \(X\) is separable.
(Hint: For (ii) \(\Rightarrow\) (iii), consider the open cover \(\{B(x, 1/n) : x \in X, n \in \mathbb{N}\}\), fix \(n\) and first use the Lindelöf property.)
Solution
Proof:
Proof of (i) \(\Rightarrow\) (ii): Second countable \(\Rightarrow\) Lindelöf
Let \(\mathcal{B}\) be a countable basis for \(X\) and let \(\mathcal{U}\) be an arbitrary open cover of \(X\).
For any point \(x \in X\), since \(\mathcal{U}\) covers \(X\), there exists an open set \(U_x \in \mathcal{U}\) such that \(x \in U_x\).
Since \(\mathcal{B}\) is a basis, there exists a basis element \(B_x \in \mathcal{B}\) such that \(x \in B_x \subseteq U_x\).
Consider the collection \(\mathcal{B}' = \{B_x \mid x \in X\}\) of all chosen basis elements. Since \(\mathcal{B}\) is countable, the subset \(\mathcal{B}'\) must also be countable.
For each basis element in \(\mathcal{B}'\), choose one open set in \(\mathcal{U}\) that contains it. The selected family is a subcollection of \(\mathcal{U}\) and is at most countable.
This countable family still covers all of \(X\). Therefore every open cover has a countable subcover, so \(X\) is Lindelöf.
Proof of (ii) \(\Rightarrow\) (iii): Lindelöf \(\Rightarrow\) separable
For any fixed positive integer \(n\), consider the family of open balls of radius \(1/n\) centered at every point of \(X\):
Clearly \(\mathcal{U}_n\) is an open cover of \(X\). Since \(X\) is Lindelöf, there exists a countable subcover.
That is, there exist countably many points \(x_1^{(n)}, x_2^{(n)}, \dots \in X\) such that the corresponding open balls cover \(X\):
Let \(D_n = \{x_i^{(n)} \mid i \in \mathbb{N}\}\) be the set of these countably many centers. \(D_n\) is countable.
Let \(D = \bigcup_{n=1}^\infty D_n\). A countable union of countable sets is still countable.
We prove that \(D\) is dense in \(X\). Take any \(y \in X\) and any \(\epsilon > 0\). There exists a positive integer \(n\) such that \(1/n < \epsilon\).
Since the balls with centers in \(D_n\) cover \(X\), there exists some \(x_k^{(n)} \in D_n \subseteq D\) such that \(y \in B(x_k^{(n)}, 1/n)\).
This means \(d(y, x_k^{(n)}) < 1/n < \epsilon\), so \(x_k^{(n)} \in B(y, \epsilon)\). Thus \(D\) intersects every open neighbourhood of \(y\). Hence \(D\) is dense in \(X\), and \(X\) is separable.
Proof of (iii) \(\Rightarrow\) (i): Separable \(\Rightarrow\) second countable
Suppose \(X\) is separable, so there exists a countable dense subset \(D\).
Construct the family of open balls:
Since \(D\) and \(\mathbb{N}\) are countable, \(\mathcal{B}\) is a countable family. We claim that \(\mathcal{B}\) is a basis for \(X\).
Take an arbitrary open set \(U\) in \(X\) and a point \(y \in U\). Since \(U\) is open, there exists \(\epsilon > 0\) such that \(B(y, \epsilon) \subseteq U\).
Choose an integer \(n\) large enough so that \(1/n < \epsilon / 2\).
Since \(D\) is dense in \(X\), there exists a point \(x \in D\) inside the open ball \(B(y, 1/n)\). That is, \(d(y, x) < 1/n\).
Now the basis element \(B(x, 1/n) \in \mathcal{B}\) contains \(y\), because \(d(y, x) < 1/n\).
For any \(z \in B(x, 1/n)\), by the triangle inequality:
Hence \(B(x, 1/n) \subseteq B(y, \epsilon) \subseteq U\). Since every open set can be approximated by elements of \(\mathcal{B}\), \(\mathcal{B}\) is indeed a countable basis for the space. Hence \(X\) is second countable.
Problem 13
Let \((X, d)\) be a pseudometric space. Let \(\mathcal{T}_d\) be the topology on \(X\) generated by the open balls
where \(B(x, \epsilon) := \{y \in X : d(x, y) < \epsilon\}\). Prove that the topological space \((X, \mathcal{T}_d)\) is a \(T_0\) space if and only if \(d\) is a metric.
Solution
Proof of necessity (\(\Rightarrow\)):
Suppose \((X, \mathcal{T}_d)\) is a \(T_0\) space. To prove that \(d\) is a metric, it suffices to prove that for all \(x \ne y\), \(d(x, y) > 0\) (since a pseudometric already satisfies symmetry, triangle inequality, and nonnegativity).
For \(x \ne y\), since the space is \(T_0\), there exists an open set \(U\) containing one of the points but not the other. Suppose, without loss of generality, that \(x \in U\) and \(y \notin U\).
Since \(\mathcal{T}_d\) is generated by open balls (in fact, the open balls themselves form a basis for the pseudometric topology), there exists an open ball \(B(p, r)\) such that:
Since \(y \notin U\), we must have \(y \notin B(p, r)\).
By the definition of open balls, \(d(p, x) < r\) and \(d(p, y) \ge r\).
Using the triangle inequality for pseudometrics, \(d(p, y) \le d(p, x) + d(x, y)\), hence:
Since \(d(x, y) > 0\) for all \(x \ne y\), \(d\) satisfies the positivity condition and is therefore a genuine metric.
Proof of sufficiency (\(\Leftarrow\)):
Suppose \(d\) is a metric. Take any two distinct points \(x \ne y\) in \(X\).
Since the metric satisfies positivity, we have \(d(x, y) = r > 0\).
Choose \(\epsilon = r / 2 > 0\) and consider the open ball \(B(x, \epsilon)\).
Clearly \(x \in B(x, \epsilon)\).
If \(y \in B(x, \epsilon)\), then by definition \(d(x, y) < \epsilon = r / 2\), contradicting \(d(x, y) = r\).
Hence \(y \notin B(x, \epsilon)\).
We have found an open set \(B(x, \epsilon)\) that contains \(x\) but not \(y\). Thus the space satisfies the \(T_0\) separation axiom.
Problem 14
For a topological space \((X, \mathcal{T})\), prove the following statements:
(i) If \(X\) is a \(T_3\) space, then for any distinct \(x, y \in X\), there exist open sets \(U, V\) such that \(x \in U\), \(y \in V\), and \(\overline{U} \cap \overline{V} = \emptyset\). (ii) If \(X\) is a \(T_4\) space, then for any disjoint closed sets \(A, B \subseteq X\), there exist open sets \(U, V\) such that \(A \subseteq U\), \(B \subseteq V\), and \(\overline{U} \cap \overline{V} = \emptyset\).
Solution
(i) Proof for \(T_3\) spaces
Suppose \(X\) is a \(T_3\) space, i.e. it is regular and satisfies the \(T_1\) separation axiom. For any distinct \(x, y \in X\), by the \(T_1\) property the singleton \(\{y\}\) is closed.
Since \(x \notin \{y\}\) and \(\{y\}\) is closed, by regularity there exist disjoint open sets \(U_1\) and \(W_1\) such that:
Since \(U_1 \cap W_1 = \emptyset\), we have \(U_1 \subseteq X \setminus W_1\). Because \(W_1\) is open, \(X \setminus W_1\) is closed.
Since \(U_1\) is contained in the closed set \(X \setminus W_1\), its closure must also be contained in it:
This means \(\overline{U_1} \cap W_1 = \emptyset\), hence \(y \notin \overline{U_1}\).
Now we have the point \(y\) and the closed set \(\overline{U_1}\) not containing \(y\). Applying regularity again, there exist disjoint open sets \(V\) and \(W_2\) such that:
Similarly, since \(V \cap W_2 = \emptyset\), we deduce \(\overline{V} \subseteq X \setminus W_2\).
Combining \(\overline{U_1} \subseteq W_2\), we obtain \(\overline{U_1} \subseteq X \setminus \overline{V}\), i.e.:
Let \(U = U_1\). We have found open sets \(U\) and \(V\) such that \(x \in U\), \(y \in V\), and \(\overline{U} \cap \overline{V} = \emptyset\).
(ii) Proof for \(T_4\) spaces
Suppose \(X\) is a \(T_4\) space, i.e. it is normal and satisfies the \(T_1\) separation axiom. Let \(A, B\) be two disjoint closed sets.
By normality, there exist disjoint open sets \(U_1\) and \(W_1\) such that:
As before, since \(U_1 \cap W_1 = \emptyset\), we have \(U_1 \subseteq X \setminus W_1\), and taking closures:
Since \(B \subseteq W_1\), we get \(\overline{U_1} \cap B = \emptyset\).
Now we have two new disjoint closed sets: \(\overline{U_1}\) and \(B\). Applying normality again, there exist disjoint open sets \(U_2\) and \(V\) such that:
Since \(U_2 \cap V = \emptyset\), similarly we deduce \(\overline{V} \subseteq X \setminus U_2\).
Because \(\overline{U_1} \subseteq U_2\), this immediately implies:
Let \(U = U_1\). We clearly have \(A \subseteq U\) (by step 2) and \(B \subseteq V\) (by step 5), and the closures are disjoint: \(\overline{U} \cap \overline{V} = \emptyset\). This completes the proof.
Problem 15
Prove that every locally compact Hausdorff space is regular, i.e. “locally compact + \(T_2 \Rightarrow T_3\)”.
Solution
Proof:
Let \(X\) be a locally compact Hausdorff space. Since \(X\) is Hausdorff (i.e. \(T_2\)), it satisfies the \(T_1\) separation axiom. To prove regularity, it suffices to show that for every closed set \(A \subseteq X\) and every point \(x \notin A\), there exist disjoint open sets separating them.
Since \(X\) is locally compact, the point \(x\) has a compact neighbourhood \(K\) containing \(x\).
By the definition of neighbourhood, there exists an open set \(W\) such that:
Since compact subsets of Hausdorff spaces are closed, \(K\) is closed.
Define \(C = A \cap K\). Since both \(A\) and \(K\) are closed, \(C\) is a closed subset of \(K\). A closed subset of a compact set is compact, so \(C\) is compact.
Since \(x \notin A\), clearly \(x \notin C\). Now we need to separate the point \(x\) from the compact set \(C\) in the Hausdorff space \(X\).
For each \(y \in C\), since \(x \ne y\) and \(X\) is Hausdorff, there exist disjoint open sets \(U_y\) and \(V_y\) such that:
The family \(\{V_y \mid y \in C\}\) is an open cover of the compact set \(C\). By compactness, there exist finitely many points \(y_1, y_2, \dots, y_n \in C\) such that:
Now define:
Clearly \(U\) and \(V\) are disjoint open sets and satisfy \(x \in U\) and \(C \subseteq V\).
Now use \(U, V\) to construct two new open sets \(O_1\) and \(O_2\) separating \(x\) from the original closed set \(A\):
Check that \(O_1, O_2\) are open and contain the desired sets:
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Since \(U\) and \(W\) are open, \(O_1\) is open. Also \(x \in U\) and \(x \in W\), so \(x \in O_1\).
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Since \(V\) is open and \(K\) is closed (so \(X \setminus K\) is open), \(O_2\) is open.
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For any \(a \in A\): if \(a \in K\), then \(a \in A \cap K = C\), hence \(a \in V \subseteq O_2\); if \(a \notin K\), then \(a \in X \setminus K \subseteq O_2\). In either case \(A \subseteq O_2\).
Check that \(O_1\) and \(O_2\) are disjoint:
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The first part is empty because \(U \cap V = \emptyset\), so \(U \cap W \cap V = \emptyset\).
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The second part is empty because \(W \subseteq K\), so \(W \cap (X \setminus K) = \emptyset\).
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Hence \(O_1 \cap O_2 = \emptyset\).
In summary, we have found two disjoint open sets \(O_1\) and \(O_2\) containing \(x\) and the closed set \(A\) respectively. Therefore the space is regular (\(T_3\)).