# Unit ball in $\mathbb{R}^{n}$ is not compact

Question:

Prove that the open unit ball $B_{1}\left ( \vec{0} \right )$ in $\mathbb{R}^{n}$ is not compact.

I have the definition of "compact" but I am unable to start on this question. Taking a fresh look at my notes, I suspect this has to do with the Heine-Borel theorem.

Any help is appreciated.

• It's not closed and $\Bbb R^n$ is $T2$. – user228113 Oct 13 '16 at 15:00

There is no finite subcover of $\{\,B_{1-\frac1n}(\vec0)\mid n\in\Bbb N\,\}$

• I too was trying to come up with some infinite covering not having a finite subcovering, but my tries were much more fancy. The simplicity of your covering is beautiful (+1). And it's actually extremely easy to generalize it to any open bounded subset, say $\Omega$, by picking the familiy of sets : $$A_n:=\{x\in \Omega\mid d(x,\partial \Omega)>1/n\}$$ – b00n heT Oct 13 '16 at 15:08
• Giving you an upvote for the simplicity. – Mathematicing Oct 13 '16 at 15:25
• @Mathematicing I was wondering if $\{\,B_r(\vec0)\mid 0<r<1\,\}$ would be simpler ... – Hagen von Eitzen Oct 13 '16 at 15:34
• @HagenvonEitzen Allow me to try that tomorrow. This could be an alternate demonstration of a proof. – Mathematicing Oct 13 '16 at 15:50

This is basically an automatic consequence of the Heine Borel theorem (which as stated in your question, you already have):

Theorem: a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.

• I have $B_{1}\left ( \vec{0} \right )= \left ( -\epsilon ,+\epsilon \right ) \subseteq \mathbb{R}^{n}$ which is not closed. I misused my language in the earlier comment. – Mathematicing Oct 13 '16 at 15:49