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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
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3 Answers

up vote 6 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
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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
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If you are interested, Spacebook provides some more examples of metrizable spaces that are not locally compact.

Baire Product Metric on Countable Cartesian Product of Reals
Cantor’s Leaky Tent
Cantor’s Teepee
Continuous Functions on the Unit Interval
Countable Cartesian Product of Positive Integers
Discrete Rational Extension of the Reals
Duncan’s Space
Evenly Space Integer Topology
Hilbert Space
Metrizable Tangent Disc Topology
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

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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
    
@Paul It cannot be downloaded, but all the data in Spacebook (and much more) can be found in Steen and Seebach's book Counterexamples in Topology. –  Austin Mohr Aug 2 '12 at 17:21
    
Okey. Thank you all the same for your answer. –  Paul Aug 2 '12 at 23:56
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