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Which of the following subsets of $\mathbb{R}^2$ are compact?

  • (a) $\left\{ (x, y) : xy = 1 \right\} $
  • (b) $\left\{ (x, y) : x^{2/3} + y^{2/3} = 1 \right\}$
  • (c) $\left\{(x, y) : x^2 + y^2 < 1\right\}$

clearly a and c are not compact. not sure about b

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Do you know the characterization of compact subsets of $\Bbb R^n$? –  Andrea Mori Sep 1 '12 at 18:02
    
Besides, this is homework, isn't it? (if it is, you're supposed to tag it as such) –  Andrea Mori Sep 1 '12 at 18:03
    
en.wikipedia.org/wiki/Astroid –  a.r. Sep 1 '12 at 18:09
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2 Answers 2

up vote 3 down vote accepted

Is the set bounded? For all $x\in\Bbb R$, $x^{2/3}\ge 0$, so if $x^{2/3}+y^{2/3}=1$, how big can $x$ and $y$ be?

Is it closed? That’s harder to answer rigorously, but a glance at the graph of the expression should give you a pretty good idea.

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$f:(x,y)\mapsto x^{2/3}+y^{2/3}$ is continuous so $f^{-1}(1)$ is closed. –  Sriti Mallick Aug 21 '13 at 2:02
    
@Sriti: Yes, if one knows that fact about continuous functions, there’s an easy way to justify the conclusion. –  Brian M. Scott Aug 21 '13 at 5:03
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A subset of $\mathbb{R}^n$ is compact iff it is closed and bounded.

a) closed but unbounded so not compact.

b) closed and bounded.see the graph

c) Open set.

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