1
$\begingroup$

We know that if $A$ and $B$ are compact (assuming A and B are non-empty), then the Cartesian product $A \text{x} B$ is compact. But how do you go the other way round.

We have to show that any sequence $(a_k)$ in A and $(b_k)$ in B have subsequences that converges in A and B respectively. We are given that any subseqence of the sequence $(a_k, b_k)$ is convergent in $A \text{ x } B$. I have at loss at how to I use this information to claim that subsequences of $(a_k) \text{and} (b_k)$ are convergent in $A$ and $B$ respectively.

Should I claim that $(a_k, b_k)$ is convergent iff each $a_k$ and $b_k$ is convergent, and be done with it?

  • 1
    An easier strategy might be to pick a point $a\in A$ and show that $\{a\}\times B$ is closed in $A\times B$. Since a closed subspace of a compact space is compact, you're done. (Noting that $B$ is clearly homeomorphic to $\{a\}\times B$.) A similar argument gives you that $A$ is compact.2012-10-31
  • 0
    Shawn, can we make the same argument without stating the last point (B is homeomorphic to {a} x B)? I am just trying to keep it really basic.2012-10-31
  • 1
    Do you mean compact _metric_ spaces? In general sequential compactness is not the same as compactness. Shawn's suggestion is the easiest proof for general compact _Hausdorff_ spaces. The easiest proof for general compact _topological_ spaces is to use projections and the fact that the image of a compact space is again compact.2012-10-31
  • 0
    @ZhenLin: am sorry, I should have clarified. Yes, I am talking about sequential compactness.2012-11-01

1 Answers 1

2

HINT: Let $\langle a_k:k\in\Bbb N\rangle$ be a sequence in $A$. Let $b$ be any point of $B$, and consider the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle$$ in $A\times B$. The same basic idea works to handle sequences in $B$.

Added: Here’s a rough sketch that may help a bit.

enter image description here

I’ve drawn this as if $A$ and $B$ were $[0,1]$, because that’s easy to visualize. The red dots in $A$ (down at the bottom) are four terms of the sequence $\langle a_k:k\in\Bbb N\rangle$; the red dots in $A\times B$ (the square) are all on the line representing $A\times\{b\}$: they are the corresponding terms of the sequence $$\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle\tag{1}$$ in $A\times B$.

You know that every sequence in $A\times B$ has a convergent subsequence, so $(1)$ converges to some point $\langle x,y\rangle\in A\times B$. In fact, $y=b$; why? Thus, $\langle x,y\rangle=\langle x,b\rangle$ is on the same ‘line’ as the sequence $(1)$, the set $A\times\{b\}$. Now show, using the definition of the product topology, that $\langle a_k:k\in\Bbb N\rangle$ converges to $x$.

  • 0
    @BrianMScott: Try fleshing that hint out for me. I was thinking of how to use the fact that the sequence $(a_k,b_k)$ in AxB can be used to arrive that the conclusion we need. I think I have a hard time thinking about the sequence $(a_k,b_k)$, ie, ordered pairs forming a sequence. I would like some help in envisioning this, and an explication will help. I have found a proof to my question, but I rather understand it before I look at it.2012-10-31
  • 0
    @user43901: You aren’t using a sequence $\big\langle\langle a_k,b_k\rangle:k\in\Bbb N\big\rangle$; you’re using a sequence $\big\langle\langle a_k,b\rangle:k\in\Bbb N\big\rangle$ whose terms all have the same second component. I’m not sure that I can say much more without just doing the proof; let me think about that while I produce a sketch that may help you to visualize what’s going on.2012-10-31
  • 0
    @user43901: I added both a sketch and a considerable further hint; see if they’re enough for you to finish the job.2012-10-31
  • 0
    @BrianMScott: thank you so much for adding this. I am working through it now, but I apologize for not understanding when you say "use the definition of product topology" since I am not aware of that. So if there is a way we can use the basic facts of compactness and convergences instead of using other terms, I would be grateful.2012-10-31
  • 0
    @user43901: Do you know that if $V\subseteq A$ and $W\subseteq B$ are open sets, then $V\times W$ is an open set in $A\times B$? And that if $U$ is an open neighborhood of a point $p\in A\times B$, there are open sets $V\subseteq A$ and $W\subseteq B$ such that $p\in V\times W\subseteq U$? I’m hard-pressed to see how you can be expected to prove the result if you don’t have at least that much information.2012-10-31
  • 0
    @BrianMScott: I think I forgot to mention that I am talking about sequentially compact. And I can see how the statements you state are true, but no, I was not aware of the statement before you put it out. All proofs I have been working on deals with sequences and subsequences converging. After researching, I see that usually compactness is defined by open covers, which is different from the definition that I am aware of2012-11-01
  • 0
    @user43901: I figured from the statement of the problem that you were working with sequential compactness; that’s why I suggested the argument using sequences. But you can’t hope to prove anything relating convergence in a product to convergence in the factors if you don’t have some basic knowledge about the product topology $-$ at least what I mentioned in the previous comment. I can’t help thinking that you’ve been turned loose on this problem with rather minimal weaponry!2012-11-01