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.

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?

share|improve this question
    
    
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

3 Answers 3

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:)

share|improve this answer
    
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.

share|improve this answer

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.

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.