Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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 at 19:27

4 Answers 4

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.

share|cite|improve this answer
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 at 12:13
Do you want a product of countable many spaces or any product? – jeanmfischer Sep 6 at 10:53
Hi. Any product. I think this theorem holds for any product, doesn't it? – Zheng Liu Sep 7 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 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 $\{ a \} \times B \subseteq U$.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.