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You have two connected topological spaces $(A,B)$. Prove that $A\times B$ is also connected.

I understand that I have to prove that there is a point in $B$ (call it $b$), that makes $A\times\{b\}$ homeomorphic to $A$ making it connected to $A\times B$. Then prove that $\{a\}\times B$ is connected in $A\times B$. But I don't really know where to being with this. If you could help that would be appreciated.

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Welcome to MSE, @moe. Please have a look at the FAQ about homework questions and take some time to learn some MathJax basics. – A.P. Mar 22 '13 at 16:36
Thank you I just started on here so I'm wasn't exactly sure how to put things up. – moe Mar 22 '13 at 17:29
One year later and still no accepted answer? Probably, you've just forget about this Q, but it is a good habit to accept answer, if they help of course. – quapka Jun 1 '15 at 19:27

Let $F : A \times B \to \{0,1\}$ be a continuous functions. To show that $A\times B$ is connected for the product topology we have to show that $F$ is constant.

As you suggested (kind of) we first show that $F$ is constant on every set of the form $\{a\}\times B$. Indeed if we have $a\in A$ we get a function $f:B \to \{0,1\}$ defined by $b \mapsto F(a,b)$. This functions is continuous thus constant because $B$ is connected.

In the exact same way we can show that $F$ is constant on the sets of the form $A \times \{b\}$.

We now show that this implies that $F$ is constant on $A\times B$. Indeed fix $(a,b) \in A \times B$. Now let's consider another point $(a',b')\in A \times B$. By what we have done earlier we have $F(a,b)=F(a,b')=F(a',b')$. We are done.

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It's a nice proof. I'm thinking how to apply this proof to show the connectivity when it is a product topology of infinitely many connected space. Any suggestions? – Zheng Liu Sep 4 '15 at 12:13
Do you want a product of countable many spaces or any product? – jeanmfischer Sep 6 '15 at 10:53
Hi. Any product. I think this theorem holds for any product, doesn't it? – Zheng Liu Sep 7 '15 at 4:42
Maybe! Let $(X_i)_{i \in I}$ be a family of topological spaces over a set $I$, and let $i \in I$. This time the $\{a\}\times B$ can be a set of the form $\{(x_j)_{j \in I -\{ i\} }\} \times X_i$, where $(x_j)_{j \in I -\{ i\} }$ lies in the product where $X_i$ was removed. – jeanmfischer Sep 7 '15 at 10:17

Suppose $U , V \subseteq A \times B$ are disjoint open sets whose union is all of $A \times B$. Fixing some $b \in B$, note that the subspace $A \times \{ b \}$ of $A \times B$ is homeomorphic to $A$, and $A \times \{ b \} \subseteq U \cup V$. By the connectedness $A$ (and hence of $A \times \{ b \}$) we may conclude, without loss of generality, that $A \times \{ b \} \subseteq U$.

Now given $a \in A$, knowing that $\langle a , b \rangle \in U$ go through a similar argument as above to conclude that $\{ a \} \times B \subseteq U$.

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Theorem. If $\{X_i\}_{i\in I}$ is a family of connected spaces such that $\bigcap_{i\in I} X_i\neq \phi$ then $\bigcup_{i\in I} X_i$ is connected.

Using this it is easy to prove what you want:

Fix $y\in Y$ and consider for every $x\in X$ the set, $$U_x=(\{x\}\times Y)\cup (X\times \{y\}).$$ Then every $U_x$ is connected for it is union of connected sets ($\{x\}\times Y\simeq Y$ and $X\times \{y\}\simeq X$) with non-empty intersection ($(\{x\}\times Y)\cap (X\times \{y\})=(x, y)$). It is easy to see $$X\times Y=\bigcup_{x\in X} U_x,$$ and since $\displaystyle \bigcap_{x\in X} U_x=X\times \{y\}\neq \phi$, $X\times Y$ is connected.

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Hint: Try to write a continuous function $f:A\times B\to \{0,1\}$ and recall that a topological space is connected if and only if every continuous function from it to $\{0,1\}$ is constant.

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