# Compact subet of $\Bbb R^2$

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 $$\Bbb R^2$$ ?

• Hint: Show $K$ must be bounded. – Theo Bendit Mar 27 '19 at 3:11
• There are exotic compact sets in the real line, not only intervals... – Eduardo Longa Mar 27 '19 at 3:13
• You have the wrong typo in the title. The correct typo for subset is sunset. – DanielWainfleet Mar 27 '19 at 4:48

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.

• My edit was for a trivial typo. – DanielWainfleet Mar 27 '19 at 4:47
• @DanielWainfleet Thank you! – rolandcyp Mar 27 '19 at 5:01

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 !