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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.

Thanks in advance.

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  • $\begingroup$ It's not closed and $\Bbb R^n$ is $T2$. $\endgroup$ – user228113 Oct 13 '16 at 15:00
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There is no finite subcover of $\{\,B_{1-\frac1n}(\vec0)\mid n\in\Bbb N\,\}$

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  • $\begingroup$ 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 : $$A_n:=\{x\in \Omega\mid d(x,\partial \Omega)>1/n\}$$ $\endgroup$ – b00n heT Oct 13 '16 at 15:08
  • $\begingroup$ Giving you an upvote for the simplicity. $\endgroup$ – Mathematicing Oct 13 '16 at 15:25
  • $\begingroup$ @Mathematicing I was wondering if $\{\,B_r(\vec0)\mid 0<r<1\,\}$ would be simpler ... $\endgroup$ – Hagen von Eitzen Oct 13 '16 at 15:34
  • $\begingroup$ @HagenvonEitzen Allow me to try that tomorrow. This could be an alternate demonstration of a proof. $\endgroup$ – Mathematicing Oct 13 '16 at 15:50
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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.

Question: Is your set closed?

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  • $\begingroup$ I started with R^{1} space. Then showed that the ball centered at a point x=0 in R^{1} is an open space; or not a closed space. This generalises to R^{n}. And the result follows from the contrapositivity of the HB theorem. $\endgroup$ – Mathematicing Oct 13 '16 at 15:23
  • $\begingroup$ @Mathematicing Open does not immediately imply not closed. $\endgroup$ – Slade Oct 13 '16 at 15:40
  • $\begingroup$ 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. $\endgroup$ – Mathematicing Oct 13 '16 at 15:49

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