# Product of connected metric spaces

I want to show that if $X$ and $Y$ are connected metric spaces, then $X \times Y$ is also connected. As a hint it is marked to write $X \times Y$ as Unions of sets of the form $(X \times \{y\}) \cup ( \{x\} \times Y)$, but i don't have an idea how to continue. Any hints?

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–  Martin Sleziak Jul 7 '12 at 3:44
Do you need to use the hint? It seems to me simpler to use the following characterization: $X$ is connected if and only if every continuous map from $X$ onto $\{0,1\}$ (with the discrete topology) is constant. –  Taladris Nov 5 '13 at 1:58

Another different proof from wiki:

Let us begin with a Lemma:

If a topological space $X$ is connected iff for any point $x \in X$ and for any open cover $U$ of $X$, we have $\cup^\omega_{n=1}St^n(x,U)=\cup \{u \in U: x\in u\}=X$, where $St^1(x,U)=\cup \{u \in U: x\in u\}$, and when $n>1$, $St^n(x,U)=St(St^{n-1}(x,U)).$

Proof:

• Left to right: We first assume that $X$ is connected. For any open cover $U$ of $X$, and for any $x\in X$, if there is a point $y$ such that $y$ is not in $\cup^\omega_{n=1}St^n(x,U)$, therefore there exists open set family $V \subset U$ $y$ such that $\cup V \cap \cup^\omega_{n=1}St^n(x,U)= \emptyset$, which implies $\cup V$ is an nonempty open-closed subspace of $X$, contradiction!
• Right to left: if the space is not connected, then exists an nonempty closed-open subspace $Y$. We let the open cover $Y\cup (X-Y)$, obviously, it can't satisfies the condition.

Then the proof for the question is not difficult, I believe. Try. (Drawing a picture will be more clear:)

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Your lemma is very similar to "chain-characterisation" of connectedness from Henno Brandsma's answer here here. –  Martin Sleziak Jul 7 '12 at 5:32
It seems same in the essence. We always call it the star-method. I like to character the connectness of the space, because it is very pictureful:) –  Paul Jul 7 '12 at 5:45

As for each $x\in X$ the subspace $\{x\}\times Y$ is homeomorphic to the connected space $Y$ (via the projection), it is connected. The same holds for all $X×\{y\}$, $y\in Y$. Now if $(x,y)$ and $(x',y')$ are points in $X×Y$, then $$(\{x\}×Y) \cup (X×\{y'\})$$ is connected since it's intersection $\{(x,y')\}$ is non-empty. This shows that there is a connected set containing $(x,y)$ and $(x',y')$. Since these points were arbitrary, $X×Y$ is connected.

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HINT: Try it with $[0,1]\times [0,1]$ first. How can you apply the hint here? Once you can do this, the general case should follow fairly quickly.

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