# Limit point compactness

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!

• Note that there are possibly subsets of $X$ that are neither open or closed. So "not closed" does not necessarily mean "open". – T. Eskin Apr 3 '14 at 1:40
• 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! – Asaf Karagila Apr 3 '14 at 1:42
• @AsafKaragila, is that a pun? – IAmNoOne Apr 3 '14 at 1:54
• @Nameless: I'll give you $42$ guesses. – Asaf Karagila Apr 3 '14 at 1:55

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.

• 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? – LittleKnown Apr 3 '14 at 1:49
• @LittleKnown $A$ is a subset of itself. – Pedro Tamaroff Apr 3 '14 at 1:50
• But A could have other limit points in itself, so it could still be limit point compact. Is it not so? – LittleKnown Apr 3 '14 at 1:57
• @LittleKnown I have edited my answer. Let me know. – Pedro Tamaroff Apr 3 '14 at 2:08
• @LittleKnown Did you try anything? – Pedro Tamaroff Apr 3 '14 at 2:54

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.

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