Proof of part of the Heine-Borel Theorem

Question

I wanna prove a portion of the Heine-Borel Theorem because I didn't like the proof I was given. I'm only going to be proving the closed part. I will not be proving the bounded part because that part was fine.

Theorem In $\mathbb{R}^n$ with the Euclidean topology, compact sets are precisely the closed and bounded subsets of $\mathbb{R}^n$

Proof

Let $K$ be a compact (bounded) subset of $\mathbb{R}^n$. Toward a contradiction, suppose $K$ is not closed.

Since $K$ is not closed, let $x_o$ be an accumulation point of $K$ which is not in $K$.

Consider all $k \in K.$ Since $K$ is bounded, let $M =$ max$\{|k_1|,|k_2|,|k_3|,...\}$. Then for all $k \in K,$ $-M \leq k \leq M,$ bounding all of $K$.

Consider $\mathcal{U}_i =\{k \in K: d(k,x_o)<\frac{M}{i}\}$ for $i \in \mathbb{N}$

Then $\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is an open cover of $K$ with no finite subcover

This contradicts our assumption that $K$ was not closed.

Thus, the compact set $K$ is closed.

Answer 1

$M$ as written is not well-defined. If $K$ is infinite, then the set $\{|k_1|, |k_2|, \dots\}$ may not have a maximum element.

There is a type error in your inequality $-M \leq k \leq M$. $M$ is a real number, but $k$ is a vector.

Similarly, there is a type error in "$\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is an open cover...". Open covers are sets of subsets, $\displaystyle\bigcup_{i=1}^\infty \mathcal{U}_i$ is just a single set.

Furthermore, the most important part, that, $\mathcal{U}_i$ is an open cover without a finite subcover is, well, wrong. Each $\mathcal{U}_i$ is smaller than the one proceding it, so a finite subcover would be given by $\mathcal{U}_1$.

Answer 2

Duncan Ramage has explained what is wrong with your argument. An argument along somewhat similar lines is possible, however, and is actually a bit simpler than what you were trying to do, since it does not require any information on whether $K$ is bounded. It begins as you did.

If $K$ is not closed, let $x_0\in\operatorname{cl}K\setminus K$, for $k\in\Bbb Z^+$ let

$$U_k=\left\{x\in\Bbb R^n:d(x,x_0)>\frac1k\right\}\,,$$

and let $\mathscr{U}=\{U_k:k\in\Bbb Z^+\}$; then $\bigcup\mathscr{U}=\bigcup_{k\ge 1}U_k=\Bbb R^n\setminus\{x_0\}$, so $\mathscr{U}$ is an open cover of $K$.

Suppose that $\{U_{k_1},\ldots,U_{k_m}\}$ is a finite subset of $\mathscr{U}$, where $k_1<\ldots<k_m$. Then $U_{k_1}\subseteq\ldots\subseteq U_{k_m}$, so $\bigcup_{i=1}^mU_{k_i}=U_{k_m}$. Since $x_0\in\operatorname{cl}K$, there is a point $x\in K$ such that $d(x_0,x)<\frac1{k_m}$; clearly

$$x\in K\setminus U_{k_m}=K\setminus\bigcup_{i=1}^mU_{k_i}\,,$$

so $\{U_{k_1},\ldots,U_{k_m}\}$ does not cover $K$. Thus, no finite subset of $\mathscr{U}$ covers $K$, and $K$ is therefore not compact.