Compact subet of $\Bbb R^2$

Question

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 $\Bbb R^2$ ?

Answer 1

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.

Answer 2

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 !