Take the 2-minute tour ×
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|improve this question
1  
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

4 Answers 4

(Fairly Complete) Hint: 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 connectedness 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 to conclude that $\{ a \} \times B \subseteq U$.

share|improve this answer

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|improve this answer
1  
Nice answer, but please have a look at the guideline on answering homework related questions. –  A.P. Mar 22 '13 at 16:55

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|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|improve this answer

Your Answer

 
discard

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.