[Math] Connectedness of a cartesian product

Question

Let $X,Y$ be topological spaces. Show that the productspace $X\times Y$ is connected $\Leftrightarrow X,Y$ are connected.

Could someone give me some pointers? I'm not looking for a full solution, but for some hints, to help me where I got stuck.

Proof:

$\Rightarrow$

Say $X$ is not connected, then there exist $U,V \not = \varnothing$ open, such that $X = U\sqcup V$.

Now consider $U\times Y$ and $V\times Y$. Both of these sets are non-empty.
Since $U, V\in \tau_X$ both sets are open. And since $U\cap V=\varnothing$ $(U\times Y) \cap (V\times Y) = \varnothing$.

Since $X\times Y = (U\times Y) \cup (V\times Y)$ $X\times Y$ would not be connected. Contradiction.

$\Leftarrow$

Say $X\times Y$ is not connected, then there exists $U,V\not = \varnothing$, open in $X\times Y$ such that $X\times Y = U\sqcup V$.

I was thinking of continuing in the way as ($\Rightarrow$), setting $U=U_X\times U_Y$ and $V=V_X\times V_Y$, and then showing something like $U = U_X\sqcup V_X$ which would lead to a contradiction.

But I don't think the existance of $U\sqcup V$ means $U$ can be written als $U_X\times U_Y$ can it? (see picture)

Answer 1

The proof of this fact needs a good idea: firstly you should fix some $x \in X$ and write $$X \times Y = \bigcup_{y \in Y} ((\{ x\} \times Y) \cup (X \times \{ y\}))$$ Secondly, you need to use twice the following proposition:

If $\{ U_{\alpha} \}_{\alpha}$ is a family of connected subspaces of a space $S$ and $\bigcap_{\alpha} U_{\alpha}$ is not empty, then $\bigcup_{\alpha} U_{\alpha}$ is a connected subspace of $S$.

You use it to show that all sets of the form $(\{ x\} \times Y) \cup (X \times \{ y\})$ are connected, and then you use it another time to show that $X \times Y$ is connected.