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

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. –  martini Jul 10 '12 at 12:54
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! –  M Turgeon Jul 10 '12 at 13:02

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 : ) ${}{}{}{}$ –  Matt N. Jul 10 '12 at 13:08
@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). –  user38268 Jul 10 '12 at 13:16
@BenjaLim: right, I missed the strict inequalities. –  tomasz Jul 10 '12 at 13:18
thank you all I have understood :) –  Une Femme Douce Jul 10 '12 at 13:45

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. –  Matt N. Jul 10 '12 at 13:14
One thing: Is the subspace topology on $D$ the same as the (Euclidean) topology on $\Bbb{R}^2$? –  user38268 Jul 10 '12 at 13:14
@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 $D \cap O = \overline{O}$. –  Matt N. Jul 10 '12 at 13:16
So I guess one could say that they are the same. –  Matt N. Jul 10 '12 at 13:17
thank you all I have understood :) –  Une Femme Douce Jul 10 '12 at 13:45

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 :) –  Une Femme Douce Jul 10 '12 at 13:46
@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. –  user38268 Jul 10 '12 at 22:45
sorry I really dont get your last comment. –  Une Femme Douce Jul 11 '12 at 6:33