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??

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.)

  • $\begingroup$ @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

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