Section Properties of Products of Topological Spaces
It is natural to ask what topological properties of the topological spaces \((X, \tau_X)\) and \((Y, \tau_Y)\) are inherited by the product \(X \times Y\text{.}\) We have studied Hausdorff, connected, and compact spaces, and we now consider those properties.
Activity 20.8.
Let \((X, \tau_X)\) and \((Y, \tau_Y)\) be Hausdorff spaces.
(a)
What will it take to prove that the space \(X \times Y\) with the product topology is Hausdorff?
(b)
Suppose that \((x_1,y_1), (x_2, y_2) \in X \times Y\text{.}\) What does the fact that \(X\) is Hausdorff tell us about \(x_1\) and \(x_2\text{?}\) What can we say about \(y_1\) and \(y_2\text{?}\)
(c)
Complete the proof of the following theorem.
Theorem 20.5.
If \((X, \tau_X)\) and \((Y, \tau_Y)\) are Hausdorff spaces, then \(X \times Y\) with the product topology is a Hausdorff space.
The proofs that a product of connected spaces is connected, that a product of path connected spaces is path connected, and that a product of compact spaces is compact are a bit more complicated. To prove that a product of two connected spaces is connected, we will use the result of
Activity 18.7 in
Chapter 18 that the union of connected subsets is connected if the intersection of the subsets is nonempty. A consequence of this result is the following.
Lemma 20.6.
Let \(X\) be a topological space, and let \(A_{\alpha}\) be a connected subset of \(X\) for all \(\alpha\) in some indexing set \(I\text{.}\) Let \(B\) be a connected subset of \(X\) such that \(A_{\alpha} \cap B \neq \emptyset\) for every \(\alpha \in I\text{.}\) Then \(B \cup \left(\bigcup_{\alpha \in I} A_{\alpha} \right)\) is connected.
Proof.
Let
\(X\) be a topological space, and let
\(A_{\alpha}\) be a connected subset of
\(X\) for all
\(\alpha\) in some indexing set
\(I\text{.}\) Let
\(B\) be a connected subset of
\(X\) such that
\(A_{\alpha} \cap B \neq \emptyset\) for every
\(\alpha \in I\text{.}\) For each
\(\alpha \in I\) let
\(B_{\alpha} = B \cup A_{\alpha}\text{.}\) Let
\(\beta \in I\text{.}\) Since
\(B \cap A_{\beta} \neq \emptyset\text{,}\) Lemma 20.6 shows that
\(B_{\beta}\) is connected. Given that
\(B\) is not empty, and
\(B \subseteq \bigcap_{\alpha \in I} B_{\alpha}\text{,}\) we see that
\(\bigcap_{\alpha \in I} B_{\alpha} \neq \emptyset\text{.}\) Lemma 20.6 allows us to conclude that
\(\bigcup_{\alpha \in I} B_{\alpha}\) is connected. But
\begin{equation*}
\bigcup_{\alpha \in I} B_{\alpha} = \bigcup_{\alpha \in I} (B \cup A_{\alpha}) = B \cup \left(\bigcup_{\alpha \in I} A_{\alpha}\right)\text{,}
\end{equation*}
and so \(B \cup \left(\bigcup_{\alpha \in I} A_{\alpha}\right)\) is connected.
We will use
Lemma 20.6 to show that a product of connected spaces is connected.
Theorem 20.7.
If \((X, \tau_X)\) and \((Y, \tau_Y)\) are connected topological spaces, then \(X \times Y\) with the product topology is a connected topological space.
Proof.
Assume
\((X, \tau_X)\) and
\((Y, \tau_Y)\) are connected topological spaces. Our approach to proving that
\(X \times Y\) is connected is to write
\(X \times Y\) as a union of two connected subspaces whose intersection is not empty. Let
\(a \in X\text{.}\) The space
\(X_a = \{a\} \times Y\) is homeomorphic to
\(Y\) via the inclusion map
\(i\) which sends
\((a,t) \in \{a\} \times Y\) to the point
\(t \in Y\text{.}\) Since
\(Y\) is connected, so is
\(X'\text{.}\) Let
\(b \in Y\text{.}\) The space
\(Y_b = X \times \{b\}\) is homeomorphic to
\(X\) via the inclusion map
\(i\) which sends
\((s,b) \in X \times \{b\}\) to the point
\(s \in X\text{.}\) Since
\(X\) is connected, so is
\(Y_b\text{.}\) (The verification of these homeomorphisms is left to the reader.) The point
\((a,b)\) is in
\(X_a \cap Y_b\text{,}\) so
\(X_a \cap Y_b \neq \emptyset\) for every
\(b \in Y\text{.}\) It follows that
\(X_a \cup \left( \bigcup_{t \in Y} Y_t \right)\) is connected by
Lemma 20.6. All that remains is to prove that
\(X_a \cup \left( \bigcup_{t \in Y} Y_t \right) = X \times Y\) and we will have demonstrated that
\(X \times Y\) is connected. The fact that
\(X_a \subseteq X \times Y\) and
\(Y_t \subseteq X \times Y\) for every
\(t \in Y\) implies that
\(X_a \cup \left( \bigcup_{t \in Y} Y_t \right) \subseteq X \times Y\text{.}\) It then remains to show that
\(X \times Y \subseteq X_a \cup \left( \bigcup_{t \in Y} Y_t \right)\text{.}\) Let
\((u,v) \in X \times Y\text{.}\) Then
\(u \in X\) and
\(v \in Y\) and
\((u,v) \in Y_v\text{.}\) Thus,
\(X \times Y \subseteq X_a \cup \left( \bigcup_{t \in Y} Y_t \right)\) and so
\(X \times Y = X_a \cup \left( \bigcup_{t \in Y} Y_t \right)\text{.}\) Therefore,
\(X \times Y\) is connected.
Once we know that a product of connected topological spaces is connected, we can extend that result to any finite number of connected spaces by induction.
Corollary 20.8.
Let \(X_{k}\) be a connected topological space for \(k\) from 1 to \(n\text{.}\) Then the product \(\Pi_{k=1}^n X_k\) is connected.
We conclude this section by demonstrating that a product of compact topological spaces is compact. It is also true that finite products of path connected and compact spaces are path connected and compact. The proofs are left to
Exercise 7 and
Exercise 8.
Theorem 20.9.
If \(X\) and \(Y\) are compact topological spaces, then \(X \times Y\) is a compact topological space under the product topology.
Proof.
Let \((X, \tau_X)\) and \((Y, \tau_Y)\) be compact topological spaces. Let \(\CC = \{O_{\alpha}\}\) be an open cover of \(X \times Y\) for \(\alpha\) in some indexing set \(I\text{.}\) Let \(a \in X\) and let \(Y_a = \{a\} \times Y\text{.}\) Since \(Y_a\) is homeomorphic to \(Y\text{,}\) we know that \(Y_a\) is compact. The collection \(\{O_{\alpha} \cap Y_a\}\) is an open cover of \(Y_a\text{,}\) and so has a finite sub-cover \(\{O_{\alpha_i}\}_{1 \leq i \leq n}\text{.}\) The set \(N_a = \bigcup_{1 \leq i \leq n} O_{\alpha_i}\) is an open set that contains \(Y_a\text{.}\) We will show that there is a neighborhood \(W_a\) of \(a\) that \(N_a\) contains the entire set \(W_a \times Y\text{.}\)
Cover the set \(Y_a\) with open sets that are contained in \(N_a\) (since \(N_a\) is open, we can intersect any open set with \(N_a\) and still have an open set). Each open set is a union of basis elements, so we can cover \(Y_a\) with basis elements \(U \times V\) that are contained in \(N_a\text{.}\) Since \(Y_a\) is compact, there is a finite collection \(U_1 \times V_1\text{,}\) \(U_2 \times V_2\text{,}\) \(\ldots\text{,}\) \(U_m \times V_m\) of basis elements contained in \(N_a\) that cover \(Y_a\text{.}\) Assume that each \(U_i \times V_i\) intersects \(Y_a\) (otherwise, we can remove that set and still have a cover). Let \(W_a = U_1 \cap U_2 \cap \cdots \cap U_m\text{.}\) Since \(a \in U_i\) for each \(i\text{,}\) we know that \(W_a\) is not empty. Each \(U_i\) is open in \(X\) and so \(W_a\) is open in \(X\text{.}\) Thus, \(W_a\) is a neighborhood of \(a\) in \(X\text{.}\) Now we demonstrate that \(W_a \times Y \subseteq \bigcup_{1 \leq i \leq m} U_i \times V_i\text{.}\) Let \((x,y) \in W_a \times Y\text{.}\) Since the collection \(\{U_i \times V_i\}_{1 \leq i \leq m}\) covers \(Y_a\text{,}\) the point \((a,y)\) is in \(U_k \times V_k\) for some \(k\) between \(1\) and \(m\text{.}\) So \(y \in Y_k\text{.}\) But \(x \in W_a = \bigcap_{1 \leq i \leq m} U_i\text{,}\) so \(x \in U_k\text{.}\) Thus, \((x,y) \in U_k \times V_k\) and we conclude that \(W_a \times Y \subseteq \bigcup_{1 \leq i \leq m} U_i \times V_i\text{.}\)
So for each \(a \in X\text{,}\) the set \(N_a\) contains a set of the form \(W_a \times Y\text{,}\) where \(W_a\) is a neighborhood of \(a\) in \(X\text{.}\) So \(W_a \times Y\) is covered by a finite sub-cover of our open cover \(\CC\) of \(X \times Y\text{.}\) The collection \(\{W_a \times Y\}_{a \in X}\) is an open cover of \(X \times Y\text{.}\) Since \(X\) is compact, the is a finite sub-cover \(W_1\text{,}\) \(W_2\text{,}\) \(\ldots\text{,}\) \(W_r\) of the open cover \(\{W_a\}_{a \in X}\) of \(X\text{.}\) is an open cover of \(X\text{.}\) It follows that the sets \(W_1 \times Y\text{,}\) \(W_2 \times Y\text{,}\) \(\ldots\text{,}\) \(W_r \times Y\) is a cover of \(X \times Y\text{.}\) For each \(i\text{,}\) the set \(W_i \times Y\) is covered by finitely many of the sets in \(\CC\text{,}\) and so the collection of these sets forms a finite sub-cover of \(X \times Y\) in \(\CC\text{.}\) Therefore, \(X \times Y\) is compact.
