Topology: An Inquiry-Based Approach

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 $X^{\prime }\text{.}$ 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 $X_a\cap Y_b\ne \mathrm{\varnothing }$ 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.

Let $(X,{\tau }_X)$ and $(Y,{\tau }_Y)$ be compact topological spaces. Let $\mathcal{C}=\{O_{\alpha }\}$ 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 $\{O_{{\alpha }_i}{\}}_{1\le i\le n}\text{.}$ The set $N_a=\bigcup _{1\le i\le n}{O}_{{\alpha }_i}$ 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 $U_1\times V_1\text{,}$ $U_2\times V_2\text{,}$ $\dots \text{,}$ $U_m\times V_m$ 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 $W_a\times Y\subseteq \bigcup _{1\le i\le m}{U}_i\times V_i\text{.}$ Let $(x,y)\in W_a\times Y\text{.}$ Since the collection $\{U_i\times V_i{\}}_{1\le i\le m}$ 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 $x\in W_a=\bigcap _{1\le i\le m}{U}_i\text{,}$ so $x\in U_k\text{.}$ Thus, $(x,y)\in U_k\times V_k$ and we conclude that $W_a\times Y\subseteq \bigcup _{1\le i\le m}{U}_i\times V_i\text{.}$

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 $\mathcal{C}$ 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 $W_1\text{,}$ $W_2\text{,}$ $\dots \text{,}$ $W_r$ of the open cover $\{W_a{\}}_{a\in X}$ of $X\text{.}$ is an open cover of $X\text{.}$ It follows that the sets $W_1\times Y\text{,}$ $W_2\times Y\text{,}$ $\dots \text{,}$ $W_r\times Y$ 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 $\mathcal{C}\text{,}$ and so the collection of these sets forms a finite sub-cover of $X\times Y$ in $\mathcal{C}\text{.}$ Therefore, $X\times Y$ is compact.