# Help understanding why a complete, totally bounded metric space implies every infinite subset has a limit point

I'm reading the following proof. Properties II and III are in my title, that a complete, totally bounded metric space implies every infinite subset has a limit point.

I have two questions near the end. Why does $d(x_n,x)\lt 2/n$? And secondly, why is $x$ a limit point of $A$? What other point in the neighborhood of $x$ is also in $A$? I don't get why they mention that $3/n\to 0$ as $n\to\infty$. How does that imply it's a limit point?

• Please give a reference when you quote text from somewhere else. Otherwise, how can anyone else look it up? – Carl Mummert Sep 25 '11 at 1:53

There’s a slight error in the proof: the claim should be that $d(x_n,x) \le 2/n$. To see this, let $\epsilon$ be any positive real number; the sequence of $x_m$’s converges to $x$, so there is a positive integer $m > n$ such that $d(x_m,x) < \epsilon$. By the triangle inequality $$d(x_n,x) \le d(x_n,x_m)+d(x_m,x) < \frac2n + \epsilon.\tag{1}$$ Thus, $$d(x_n,x) < \frac2n + \epsilon$$ for every $\epsilon > 0$, and hence $d(x_n,x) \le \dfrac2n$.

This small error doesn’t affect the next step of the argument: if $y \in B(x_n,1/n)$, then $d(x_n,y) < 1/n$, so $$d(x,y) \le d(x,x_n)+d(x_n,y) \le \frac2n + d(x_n,y) < \frac2n + \frac1n = \frac3n,$$ $y \in B(x,3/n)$, and therefore $B(x_n,1/n) \subseteq B(x,3/n)$.

Now $B(x_n,1/n)\cap A$ is infinite for each $n$, and $B(x_n,1/n) \subseteq B(x,3/n)$, so $B(x,3/n)\cap A$ is infinite for each $n$. Since $3/n\to 0$ as $n\to\infty$, for any $\epsilon > 0$ there is an $n_\epsilon$ such that $3/n_\epsilon < \epsilon$. But then $B(x,\epsilon)\cap A \supseteq B(x,3/n_\epsilon)\cap A$, which is infinite. Thus, every nbhd of $x$ contains infinitely many points of $A$.

Added: To clarify, it is in fact true that $d(x_n,x)<2/n$ for each $n$; it just doesn’t follow directly from the fact that $d(x_m,x_n) < 2/m$ when $m<n$, as the weaker inequality does. If we want the strict inequality, we can modify the argument that I gave by choosing $m>n$ so that $1/m<\epsilon/2$ and $d(x_m,x)<\epsilon/2$ and then replacing $(1)$ by $$d(x_n,x)\le d(x_n,x_m)+d(x_m,x)<\frac1n+\frac1m+\frac{\epsilon}{2}<\frac1n+\epsilon.$$ Since $\epsilon$ can be chosen arbitrarily small, we conclude that $d(x_n,x)\le 1/n$ for each $n$.

• I don't think this is an error. They had established $d(x_n,x_m) \leq 1/m+1/n \lt 2/n$ for $n \lt m$ already (with $n$ and $m$ interchanged), so if you fix $n$ and let $m \to \infty$ then you get strict inequality. – commenter Sep 25 '11 at 1:01
• @commenter: Their statement isn’t false, but it doesn’t follow from the fact that $d(x_n,x_m)<2/m$ when $m<n$, which is what the given wording suggests; I think that this was a genuine oversight on the author’s part. (It’s also a little more work to prove carefully, which is why I didn’t bother.) – Brian M. Scott Sep 25 '11 at 1:19
• Thanks Brian, I appreciate your clarification. – groops Sep 27 '11 at 22:21

I think it worth settling some terminology. This is from Topology: a first course by James R. Munkres. He identifies three versions of compactness for a topological space: (A) compactness, then (B) your property (which he calls "limit point compactness") and a milder condition (C) he calls "sequential compactness" which is that every infinite sequence of points has a convergent subsequence. He proves that (A) implies (B) implies (C). He also proves that the three conditions are equivalent for a metric space.

Finally, a well-known theorem is that a metric space is compact if and only if it is complete and totally bounded. So the strongest condition is what is usually discussed in this setting. However, (A) then implies (B).

Finally, and this is not obvious, the product of two compact topological spaces is compact, and the product of two sequentially compact spaces is sequentially compact. What is unexpected is that the product of two limit point compact spaces need not be limit point compact. However, if they are metric spaces, the result does hold then.

To address the first question (the other is addressed in the post by Brian). [Also, there's no error.]

Consider $\epsilon=1/(2n)$. Then there is some $N>0$ such that $d(x_m,x)<1/(2n)$ for all $m \geq N$. Let $m=\max\{N,2n\}$. Then $d(x_n,x) \leq d(x_n,x_m)+d(x_m,x) < [(1/n)+(1/m)]+1/(2n) \leq (1/n) + (1/2n) + (1/2n) = 2/n$.

• While the assertion in question is true, the presentation is in my view seriously flawed, in that it suggests a justification that doesn’t in fact work. – Brian M. Scott Sep 25 '11 at 1:36
• I'm not sure this is quite what you're looking for, but I believe the author just jumped from $d(x_n,x_m) < 2/n$ for $n<m$ to $d(x_n,x)<2/n$ reasoning as follows: The distance between two elements in the sequence is less than 2 over the index of the earlier appearing element therefore the distance between $x_n$ and $x$ is less than $2/n$ (since $x$ sort of appears "later" in the sequence). – Bill Cook Sep 25 '11 at 19:12
• If so, the author did indeed make a mistake, since that argument is invalid: it really does justify only $\le$. The alternative is that the author had the correct argument in mind but failed to give any indication of it, which is very bad writing in what appears to be a textbook. – Brian M. Scott Sep 25 '11 at 19:26
• I have to agree with you. This strikes me as a difficult to read proof of a fairly straightforward theorem. I'm no expert in this area, but there are certainly much better, cleaner, clearer proofs out there. – Bill Cook Sep 26 '11 at 0:18