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Please, help me prove the folowing statement:

"If $K \subset \mathbb R^{m}$ and $L \subset \mathbb R^{n}$ are both compact sets,then the cartesian product $K \times L \subset \mathbb R^{m+n}$ is also a compact set".

I must prove that $K \times L$ is closed and bounded..

Claim 1: $K \times L$ is closed.

Prove:

In fact. Let $f: X \rightarrow \mathbb R^n$ be a function defined in the subset $X \subset R^m$. $f$ is continuous if and only if the inverse image $f^{-1}(F)$ of every closed set $F \subset \mathbb R^n$ is a closed set in $X$. The projections $\pi_1:\mathbb R^m \times \mathbb R^n \rightarrow \mathbb R^m$ and $\pi_2 \mathbb R^m \times \mathbb R^n \rightarrow \mathbb R^n$, defined as $\pi_1 (x,y)=x$ and $\pi_2 (x,y)=y$ are both continuous and $F \times G = \pi_1^{-1} (F) \cap \pi_2^{-1}(G)$. Then $F \times G$ is closed.

Claim 2:$K \times L$ is bounded.

Proof:

$K,L$ are bounded implies that there exists $a,b \in \mathbb R^m$ and $c,d \in \mathbb R^n$ such that $a \le K \le b$ and $c \le L \le d$. But it doesn't follow that $(a,c) \le F \times L \le (b,d)$. How can I prove that $F \times L$ is bounded?

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  • $\begingroup$ "bounded", not limited. $\endgroup$ – DonAntonio Jun 17 '14 at 19:52
  • $\begingroup$ Why doesn't it follow that $(a,c)\le K\times L\le (b,d)$? $\endgroup$ – user21467 Jun 17 '14 at 19:53
  • $\begingroup$ I don't think there is an order relation in $\Bbb{R}^m$ (at least not one that comes to the for here). What do you mean by $a\le K$, when $m>1$? $\endgroup$ – Jyrki Lahtonen Jun 17 '14 at 19:55
  • $\begingroup$ It is possible to establish an order relation in $\mathbb R^m$ (lexicographic,for example). But it doesn't follow from the hypothesis). Remember that $a \in \mathbb R^m$. $\endgroup$ – Walter r Jun 17 '14 at 21:09
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Maybe you can also argue using sequences: A metric space is compact iff every sequence has a convergent subsequence. So, assume $K,L$ are both compact, so that every sequence $k_n$ in K, resp. $l_n$ in L has a convergent subsequence. Let $S_n:=(a_n,b_n)$ be a sequence in $K \times L$ . I think it follows pretty straightforward that this has a convergent subsequence $s_n:=(a_{n_m}, b_{n_m})$.

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  • $\begingroup$ Convincent argument. Thanks! $\endgroup$ – Walter r Jun 17 '14 at 21:12
  • $\begingroup$ Glad it helped, Walter. $\endgroup$ – user99680 Jun 17 '14 at 21:25

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