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If $K$ is a compact subset of $\Bbb R^2$ then prove that $K\subset [a,b]\times [c,d]$ for some pair of compact intervals $[a,b]$ and $[c,d]$.

How can I prove this? Any hint? We know that any compact set of $\Bbb R$ is of the form $[a,b]$ or any finite set. But how can I figure out the subset of $\Bbb R^2$ ?

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    $\begingroup$ Hint: Show $K$ must be bounded. $\endgroup$ – Theo Bendit Mar 27 '19 at 3:11
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    $\begingroup$ There are exotic compact sets in the real line, not only intervals... $\endgroup$ – Eduardo Longa Mar 27 '19 at 3:13
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    $\begingroup$ You have the wrong typo in the title. The correct typo for subset is sunset. $\endgroup$ – DanielWainfleet Mar 27 '19 at 4:48
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Consider the family $\left\{ G_n \right\}_{n \geqslant 1}$ where each $G_n$ is the open subset of $\mathbb{R}^2$ given by $$ G_n := (-n,n) \times (-n,n). $$ Clearly, $G_n \subset G_{n+1}$ for each $n$ and these form an open cover of $\mathbb{R}^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N \in \mathbb{N}$ such that $K \subseteq G_N$. In this case, $$ K \subseteq G_n \subset [-N,N] \times [-N,N]. $$


Additional Note. As pointed out in the comments, be warned that not every compact subset of $\mathbb{R}$ is an interval. By the Heine-Borel theorem, a set $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $\mathfrak{C}$ is compact but is neither countable nor an interval.

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    $\begingroup$ My edit was for a trivial typo. $\endgroup$ – DanielWainfleet Mar 27 '19 at 4:47
  • $\begingroup$ @DanielWainfleet Thank you! $\endgroup$ – rolandcyp Mar 27 '19 at 5:01
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There is $c>0$ such that $\sqrt{x^2+y^2} \le c$ for all $(x,y) \in K.$

Now let $(x,y) \in K.$.

Then $|x|=\sqrt{x^2} \le \sqrt{x^2+y^2} \le c$ and $|y|=\sqrt{y^2} \le \sqrt{x^2+y^2} \le c$.

Thus $(x,y) \in [-c,c] \times [-c,c].$ This gives $K \subseteq [-c,c] \times [-c,c].$

Remark: all we need is that $K$ is bounded !

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