2
$\begingroup$

Let $(X, d)$ be a metric space and $A ⊆ X$. If $A$ is limit point compact, show that $A$ is closed.

My thoughts: The definition of limit point compactness is that for each infinite subset of $A$, it contains at least one limit point of $A$.

Then I attempt to show the contrapositive: if $A$ is not closed (it could be neither open or closed or both, thanks to the suggestions), then it cannot be limit point compact. This means there is some infinite subset of $A$ that does not contain any limit point of $A$.

Can someone suggest how to proceed afterwards? Thank you!

$\endgroup$
  • 3
    $\begingroup$ Note that there are possibly subsets of $X$ that are neither open or closed. So "not closed" does not necessarily mean "open". $\endgroup$ – T. Eskin Apr 3 '14 at 1:40
  • 2
    $\begingroup$ Sets are not like Windows, they are not either open or closed; they are more like Unix, they can be open or closed or both or neither! $\endgroup$ – Asaf Karagila Apr 3 '14 at 1:42
  • $\begingroup$ @AsafKaragila, is that a pun? $\endgroup$ – IAmNoOne Apr 3 '14 at 1:54
  • $\begingroup$ @Nameless: I'll give you $42$ guesses. $\endgroup$ – Asaf Karagila Apr 3 '14 at 1:55
2
$\begingroup$

Suppose that $A$ is not closed. In particular $A$ is infinite, since finite sets in metric spaces are closed. Since $A$ is not closed, there is a a point $x\in M$ that is a limit point of $A$, but is not contained in $A$. This means there exist an infinite subset $x_i\neq x_j$ if $i\neq j$, $S=\{x_1,x_2,\ldots\}$ of $A$ such that $x_n\to x$ but $x\notin A$. I claim that $x$ is the only limit point of $S$. Can you prove this? (You have two options, either prove that if $y$ is a limit point of $S$; then $x=y$, or that if $x\neq y$; then $y$ is not a limit point of $S$).

Note that if $y$ was a limit point of $S$, for each $\varepsilon >0$ there'd exist infinitely many $x_i\in d(y,\varepsilon)$. I claim we can construct a subsequence of $(x_n)_{n\geqslant 1}$ that converges to $y$. Indeed, taking $\varepsilon=1$ there exists a least $n_1$ such that $x_{n_1}\in B(y,1)$. Removing the finite subset $S_1=\{x_1,\ldots,x_{n_1}\}$ from $S$ doesn't affect its limits points. Thus, we can find a least $n_2$ such that $x_{n_2}\in B(y,1/2)\cap (S-S_1)$. Note that since $x_{n_2}\in B(y,1)$, $n_1<n_2$. Continuing, we produce a sequence $n_1<n_2<n_3<\cdots$ for which $d(y,x_{n_k})<k^{-1}$. Evidently then $x_{n_k}\to y$. Since $x_n\to x$, $x_{n_k}\to x$ too, so $x=y$, as we wanted.

$\endgroup$
  • $\begingroup$ Does that mean we can construct an infinite subset of A that only contains x as the limit point, so A is not limit point compact (a contradiction)? If so how would such infinite subset be constructed? $\endgroup$ – LittleKnown Apr 3 '14 at 1:49
  • $\begingroup$ @LittleKnown $A$ is a subset of itself. $\endgroup$ – Pedro Tamaroff Apr 3 '14 at 1:50
  • $\begingroup$ But A could have other limit points in itself, so it could still be limit point compact. Is it not so? $\endgroup$ – LittleKnown Apr 3 '14 at 1:57
  • $\begingroup$ @LittleKnown I have edited my answer. Let me know. $\endgroup$ – Pedro Tamaroff Apr 3 '14 at 2:08
  • $\begingroup$ @LittleKnown Did you try anything? $\endgroup$ – Pedro Tamaroff Apr 3 '14 at 2:54
1
$\begingroup$

Assume that $X-A$ is not open (not $A$ is open!).

There is a $x\in X-A$ such as for every $r>0$ there is $y\in A, d(x,y)<r$.

Take $r=1/n$, $y_n$ being the correspondant $y$. Use compactness: there is a subsequence $y'_n$ such as $$ y'_n\to y\in A $$ But also $$ d(x,y_n)<\frac 1n \implies y'_n\to x $$ then $x=y\in A$, (in a metric space the limit is unique), this is impossible.

$\endgroup$
  • $\begingroup$ Sorry we used Rudin and have not yet covered sequence, so I don't quite understand the solution, but thank you for the help! $\endgroup$ – LittleKnown Apr 3 '14 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.