4
$\begingroup$

If $A_i$ is a compact subset of a metric space $(X_i,d)$ where $i = 1,2$ to show that $A_1 \times A_2$ is compact in $X_1 \times X_2$.

Proof: Let $\{(a_n ,b_n)\}$ be any sequence in $A_1 \times A_2$ . Then $a_n $ in $A_1$ has a convergent subsequence $a_{n_i} \to a$ and taking the sequence $b_{n_i}$ in $A_2$ we have a convergent subsequence $b_{n_{i_{j}}} \to b$ and also $a_{n_{i_{j}}} \to a$. Thus the sequence $\{(a_n ,b_n)\}$ have a convergent subsequence $(a_{n_{i_{j}}},b_{n_{i_{j}}} ) \to (a,b)$. Thus $A_1 \times A_2$ is compact in $X_1 \times X_2$.

Is my working correct??

$\endgroup$
  • $\begingroup$ Well, you proved that it's sequentially compact. But in a metric space, sequential compactness is equivalent to compactness, so you are ok. I think your work looks good. Let's wait to see if others agree. $\endgroup$ – layman Mar 16 '15 at 13:10
1
$\begingroup$

Little careful regarding the suffixes.

If $\{a_{n_i}\}$ is a convergent subsequence of $\{a_{n}\},$ but $\{b_{n_i}\}$ need not be a convergent subsequence of $\{b_n\}.$ Therefore, we shall look for the convergent subsequence of $\{b_{n_i}\}$ say $\{b_{n_{i_k}}\},$ then finally the subsequence $$\left\{\left(a_{n_{i_k}}, b_{n_{i_k}}\right)\right\} $$ of $\{(a_n,b_n)\}$ will do the job.

$\endgroup$
  • $\begingroup$ I'm really confused by your post. What is wrong with the OP's suffixes? $\endgroup$ – layman Mar 16 '15 at 17:00
  • $\begingroup$ @user46944 I mean if $a_{n_1},a_{n_2},a_{n_3}\cdots,$ is a convergent subsequence of $\{a_n\},$ then the subsequence of $\{b_n\}$ with suffixes $n_1,n_2,n_3\cdots$ that is $b_{n_1},b_{n_2},b_{n_3},\cdots$ need not be convergent. $\endgroup$ – Suhail Mar 16 '15 at 17:52
  • $\begingroup$ Where did their answer/notation suggest that $b_{n_{1}}, b_{n_{2}}, b_{n_{3}}, \dots$ might be convergent? $\endgroup$ – layman Mar 16 '15 at 18:27
  • $\begingroup$ @user46944 Oh sorry, I have read the given answer incorrectly, I read $i$ for $i_j.$ I must apologise, and thanks for correcting me. $\endgroup$ – Suhail Mar 16 '15 at 18:48
  • $\begingroup$ Lol I thought I was going crazy looking for the issue in his question! $\endgroup$ – layman Mar 16 '15 at 18:58
0
$\begingroup$

Yes, it looks good to me!


If you want to work from the open set definition of compactness, you might reason like this. Let $\mathcal{U}$ be an open cover of $X\times Y$. For each $x\in X$, by the compactness of $Y$ there is a finite subcover $\mathcal{U}_x$ of $\mathcal{U}$ which covers $\{x\}\times Y$. For each $x\in X$, let $U_x\subset X$ be the intersection of the projections of the open sets in $\mathcal{U}_x$.

Note that by construction the fiber of the projection map over $U_x$ is contained in the union of $\mathcal{U}_x$.

Now by compactness of $X$, we may choose finitely many $x_1,\ldots,x_n\in X$ such that $\cup_k U_{x_k} = X$. The collection $\cup_k \mathcal{U}_{x_k}$ is now a finite subcover of $X\times Y$. This is because for any $(x,y)\in X\times Y$, $x\in U_{x_k}$ for some $k$. Since $U_{x_k}$ is the intersection of projections, the set $\{x\}\times Y$ is contained in the union of $\mathcal{U}_{x_k}$, and so $(x,y)$ is contained in one of the open sets of $\mathcal{U}_{x_k}.$

(This is theorem 26.7 of Munkres. Thanks to Brian M. Scott for catching my sloppy answer and forcing me to revisit point-set more carefully.)

$\endgroup$
  • $\begingroup$ @BrianM.Scott My life is a lie! ... er, I mean, does the answer look better now? $\endgroup$ – Neal Mar 17 '15 at 19:01
  • $\begingroup$ Much better. Alternatively, you can avoid using the projection maps by refining $\mathscr{U}$ to open boxes. $\endgroup$ – Brian M. Scott Mar 18 '15 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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