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[NBHM_2006_PhD Screening Test_Topology]

which of the spaces are Locally Compact

  1. $A=\{(x,y): x,y \text{ odd integers}\}$

  2. $B=\{(x,y): x,y\text{ irrationals}\}$

  3. $C=\{(x,y): 0\le x<1, 0

  4. $D=\{(x,y): x^2+103xy+7y^2>5\}$

A topological space $X$ is locally compact if every point has a neighborhood which is contained in a compact set.

well, I can prove that $\mathbb{Q}$ is not locally compact, so 1,2, are not Locally Compact, 3 is clearly locally compact. I am not ssure about 4. thank you.

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    1. is locally compact, as it is discrete.2012-07-10
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    There is some information missing; $A$, $B$, and so on, are presumably subsets of $\mathbb{R}^2$? We need a topology on a set to decide whether or not it is locally compact!2012-07-10

3 Answers 3

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A subset of $\mathbf R^2$ is compact iff it is closed and bounded (by Heine-Borel theorem), so a subspace of $\mathbf R^2$ is locally compact iff a small enough closed ball around any given point is still closed as a subset of $\mathbf R^2$ (because compactness is absolute, and of course it is bounded). This should be enough to solve the problem by yourself.

As for the answers, 1 is locally compact as martini said, 2 is indeed not locally compact (but it does not follow from the fact $\mathbf Q$ is not locally compact), 3 is locally compact, and 4. is locally compact.

As an additional hint for 4.: notice that it is an open subset of $\mathbf R^2$.

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    Ok : ) ${}{}{}{}$2012-07-10
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    @tomasz Are you sure that 3 is compact? I think it is not even closed ( the bottom and right edges of the square are not included).2012-07-10
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    @BenjaLim: right, I missed the strict inequalities.2012-07-10
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    thank you all I have understood :)2012-07-10
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For 4):

All open or closed subsets of a locally compact Hausdorff space are locally compact in the subspace topology. $R^2$ is locally compact and Hausdorff and $D = p^{-1}((5, \infty))$ is the inverse image of an open set under a continuous function $p(x,y) = x^2+103xy+7y^2$.

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    Assuming the subspace topology.2012-07-10
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    One thing: Is the subspace topology on $D$ the same as the (Euclidean) topology on $\Bbb{R}^2$?2012-07-10
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    @BenjaLim I'm not sure I understand your question. A set $\overline{O}$ in $D$ is open in the subspace topology if and only if there is an open set $O$ in $R^2$ such that $D \cap O = \overline{O}$.2012-07-10
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    So I guess one could say that they are the same.2012-07-10
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    thank you all I have understood :)2012-07-10
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I believe 4 is locally compact when you consider $\Bbb{R}^2$ with the Euclidean topology. If you plot the region $D$ in wolframalpha, you should see why.

By the way the fact that 2) is not locally compact does not follow from $\Bbb{Q}$ being not locally compact, although the proof is similar.

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    thank you all I have understood :)2012-07-10
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    @Patience The proof of $\Bbb{Q}$ being not locally compact that I'm referring to is not the one that uses the Baire Category Theorem.2012-07-10
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    sorry I really dont get your last comment.2012-07-11