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My text book said:

Not every metrizable space is locally compact.

And it lists a counterexample as following: The subspace $Q=\{r: r=\frac pq; p,q \in Z\}$ of $R$ with usual topology, i.e., $Q$ is the set of all rationals. It said: for any open ball of any point $r \in Q$, the closure is not compact. I can't understand this sentence. Why the closures of the open balls are not compact.

Could anybody help me to understand this sentence. Thanks ahead:)

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Sorry, I deleted the comment. The point is that in a compact metric space, every sequence has a convergent subsequence. Also, every subsequence of a convergent sequence converges to the same point. So for every sequence converging to an irrational, every subsequence would converge to the same irrational in $\mathbb{R}$ which is of course not rational, so no subsequence can converge in $\mathbb{Q}$. –  Michael Greinecker Jul 31 '12 at 8:52
    
You have sequences that lie entirely in $\mathbb{Q}$ but converge to irrationals in $\mathbb{R}$. By taking a closure of a set in the relative topology of $\mathbb{Q}$ you at most gain more rationals to the set, not these limits for example. For this reason, from the closure of every open ball in $\mathbb{Q}$ you find sequences with no convergent subsequences. –  Thomas E. Jul 31 '12 at 9:02
    
@ThomasE. Sorry, I try to understand your comment, however, I can't catch you. Do you talk about the sequential compactness? (But I want to show any closure of open ball in $Q$ is not compact:) –  Paul Jul 31 '12 at 9:15
    
@Paul. Since in metric spaces compactness is equivalent with sequential compactness then it suffices. Take any open ball, it contains Cauchy sequences that have a limit in $\mathbb{R}$ but not in $\mathbb{Q}$ (a sequence that converges to an irrational number), then by closuring this open ball in $\mathbb{Q}$ you at most obtain more rationals in the set and not these limits a priori. Thus the closure of every open ball has sequences with no convergent subsequences and is hence non-compact. –  Thomas E. Jul 31 '12 at 9:52
    
@ThomasE. O, OH, yes. They are equal in the metric spaces. You are right:) –  Paul Aug 1 '12 at 6:53

3 Answers 3

up vote 7 down vote accepted

Let $I$ be any nontrivial interval, and $r$ be an irrational number in $I$. We want to show that $I\cap\mathbb{Q}$ is not compact in $\mathbb{Q}$. Let $O_n=(-\infty,r-1/n)\cap\mathbb{Q}$ and $U_n=(r+1/n,\infty)\cap\mathbb{Q}$. Show that $$\{O_n:n\in\mathbb{N}\}\cup\{U_n:n\in\mathbb{N}\}$$ is an open cover without finite subcover.

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Thanks for the answer. By your previous comment, I find another method: For any open ball, we choose a sequence from its closure which converges to an irrational, so that the closure has a infinte closed discrete subset, which contradiction with compactness. Does it OK? –  Paul Jul 31 '12 at 9:03
    
Yes, that works too. –  Michael Greinecker Jul 31 '12 at 9:05

Another possibility: A topological space is locally compact iff it is open in its compactifications (a compactification of a topological space $X$ is a compact topological space $Y \supset X$ such that $X$ is dense in $Y$).

But $\overline{\mathbb{R}}$ is a compactification of $\mathbb{Q}$ and $\mathbb{Q}$ is not open in $\overline{\mathbb{R}}$ (in fact, the interior of $\mathbb{Q}$ is empty).

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Your answer is very interesting. Very appreciated! Could you let me know where is the claim from or give me a link? –  Paul Jul 31 '12 at 9:25

If you are interested, $\pi$-Base (an online version of Steen and Seebach's Counterexamples in Topology) provides some more examples of metrizable spaces that are not locally compact. You can view the search result to learn more about these spaces.

Baire Space

$C[0,1]$

Cantor's Leaky Tent

Cantor's Teepee

Discrete Rational Extension of the Reals

Duncan's Space

Evenly Spaced Integer Topology

Hilbert Space

Metrizable Tangent Disc Topology

Miller's Biconnected Set

Nested Rectangles

The Irrational Numbers

The p-adic Topology on the Integers

The Post Office Metric

The Radial Metric

The Rational Numbers

Topologist's Sine Curve

Wheel Without Its Hub

$\mathbb{Z}^\mathbb{Z}$

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Your Spacebook is very useful and very interesting for me. How could I download it? –  Paul Aug 2 '12 at 6:51
    
Okey. Thank you all the same for your answer. –  Paul Aug 2 '12 at 23:56
    
Spacebook has been supplanted by $\pi$-Base. –  Austin Mohr Jul 29 at 21:59

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