Show that $\mathbb{R^{n}}$ is not compact by definition

Question

I want to show by definition that $\mathbb{R^{n}}$ is not compact, so I did this:

Let ${\Omega}$ be an open cover of $\mathbb{R^n}$ and suppose that $\mathbb{R^n}$ is compact, then there exists a finite subcollection of ${\Omega}_{i=1}^{k}$ such that $\mathbb{R^{n}}$ is covered by $\Omega_{i=1}^{k}$. So, ${\cup}_{i=1}^{k}\Omega_{i}$ $\supseteq \mathbb{R^{n}}$.

From this part, I don't know how to continue the proof, I think the last argument is a contradiction but I'm not quite sure. Can you help me? Please.

Answer 1

Hint: If you want to show that $\mathbb{R}^{n}$ is not compact, you just need to provide a specific example of an open cover that has no finite subcover.

Try with the open cover $\{\Omega_{k}\}_{k=1}^{\infty}$, where $\Omega_{k}$ is the open ball of radius $k$, centred at $\mathbf{0}$.

(By the way, this is an example of the general principle that to show that something is not always true, giving a specific counterexample is sufficient.)

Answer 2

How about the open cover consisting of the open balls of radius $1$: $\mathscr C=\{B(r,1)\mid r\in\Bbb R^n\}$?

I don't think there's a finite subcover.

For suppose we have a finite subcollection: $\mathscr F=\{B(r_k,1)\mid k=1,\dots,n\}$.

Take a point $x\in\Bbb R^n$ with distance from the origin $d(x,0)\gt\max \{1+d(r_k,0)\mid k=1,\dots,n\}$. You can do this because $\Bbb R^n$ is unbounded. (I guess technically you need the Archimedean principle for this.)

Then $x\not\in B(r_k,1)$ for any $B(r_k,1)\in\mathscr F$.