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I cannot solve this:

Let $X$ be a complete metric space. Suppose that for any $r> 0$ there are finite points $x_1, x_2, \dots ,x_n$ such that $N_r(x_1),\dots, N_r(x_n)$ cover $X$. Show that $X$ is compact.

egreg
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user143299
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  • It is not trivial, and it is called total boundedness. To begin, read http://en.wikipedia.org/wiki/Totally_bounded_space#Relationships_with_compactness_and_completeness – Siminore Apr 15 '14 at 13:05
  • I reformatted using LaTeX syntax. The property you mention is called “total boundedness” (or also “precompactness”) and together with completeness it indeed implies compactness. – egreg Apr 15 '14 at 13:05
  • Thanks you, egreg for converting that into nice version. I read wikipedia link in the comment. So, How can I construct Cauchy sequence using the finite points above? – user143299 Apr 15 '14 at 13:13
  • Hmmmm it looks quite difficult..... I am now in rudin's chapter 3, and this was one of 6 problems in 75 minutes exam in the lecture of my professor's last semester. I rather take exam freely because everybody cannot solve the probl like this uuu – user143299 Apr 15 '14 at 13:31
  • By "finite points", do you mean finitely many points? If not, then what is a "finite point"? – Michael Hardy Apr 15 '14 at 15:35
  • "fintely many points" I think. – user143299 Apr 16 '14 at 04:00

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Let $a_k$ be a sequence in $X$. Show that $a_k$ has a cauchy subsequence, which by completeness, converges. Then since $a_k$ is arbitrary, $X$ is sequentially compact, so compact.

Take $r_j$ decreasing to $0$. Then for $r_1$ cover $X$ by finitely many balls of radius $r_1$. There are therefore infinitely many $a_k$ in one such ball. Reduce the sequence $a_k$ to the subsequence containing only those terms in the chosen ball, and choose one such term to be $b_1$.

Continue in this way with $r_2$, $r_3$, etc to get a cauchy subsequence $b_i$ of $a_k$, where at the $n$-th step we take a cover by finitely many balls of radius $r_n$, note that one such ball has infinitely many terms of the subsequence of $a_k$ that we got from the previous step, further reduce this subsequence of $a_k$ to just the terms in this chosen ball, and then choose $b_n$ to be one such term with index chosen higher than previous terms of $b_i$. This gives a cauchy subsequence of $a_k$.

Seth
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  • Thank you for answering. Because I don't know about sequential compactness, I cannot fully understand the argument above. I would rather study about sequential compactness more, and then reading this again may help me solve this problem completely. Thanks! – user143299 Apr 16 '14 at 04:02
  • @user143299 no problem. Sequential compactness just means that every sequence has a convergent subsequence and is equivalent to compactess in metric spaces. You should be able to find a proof in most analysis or topology books. – Seth Apr 16 '14 at 12:14