I am working on a problem discussed in this link, in which it is allegedly proven that limit point compactness and countable compactness are equivalent when the space is $T_1$ Here is the relevant part, which is a proof that limit point compactness implies countable compactness:

Suppose that X is not countably compact. So there is a countable open cover (U_n)_{n in N}, without a finite subcover. Pick, for each i, x_i in X \ (U_1 / U_2 / ... / U_i); this is possible because {U_1,....,U_i} is not a cover of X by assumption. Let A be the set of these x_i (i in N). Suppose now that x is in X. Then x is in some U_N (as the (U_n) are a cover) and this U_N can only contain x_i for i < N, by definition (x_i is NOT in U_j if j <= i). So U_N is a neighbourhood that only intersects at most finitely many points of A, and so x is not a limit point of A. This holds for any x, so A does not have any limit points. This contradicts 2. So X is countably compact.

Henno's proof seems to overlook the case in which $A$ is finite; it seems that he should have begun his proof differently, for in that case, we do in fact have a finite cover. I think his proof should have begun like this:

Let $x_n$ be defined as that point in $X - U_1~ \cup ... \cup ~U_n$, should it exist, and let $A$ denote the collection of these points. If $A$ is finite, say it equals $\{x_1,...,x_n\}$, then this means that there is no point in $X- U_n~ \cup ... \cup~ U_{n+1}$, which means we found a finite cover.

At this point we suppose that $A$ is infinite and use Henno's proof to deal with that case. However, I still take issue with his proof, mostly with this:

So U_N is a neighbourhood that only intersects at most finitely many points of A, and so x is not a limit point of A.

I don't see how this proves that $A$ has no limit points. According to my book, $x$ is a limit point of $A$ if every neighborhood of $x$ intersects $A \setminus \{x\}$. There is no stipulation that it intersects infinitely many points of $A$. It seems that we need to prove the stronger claim that $U_N$ does not intersect $A$ at all, but I don't see how that is possible at present.

  • $\begingroup$ Why the downvote? Please offer justification. $\endgroup$ – user193319 Jun 21 '17 at 14:42
  • $\begingroup$ Does this answer your question? $\endgroup$ – Mikhail Katz Jun 21 '17 at 14:50
  • $\begingroup$ @MikhailKatz I am not entirely sure. In the problem give in the link, the space is assumed to be metrizable; but I am only assuming that it is $T_1$. $\endgroup$ – user193319 Jun 21 '17 at 14:58
  • $\begingroup$ $A$ is always infinite. Suppose it is not. Then there is some $p \in X$ such that for some infinite set of indices $B \subseteq \mathbb{N}$ we $p = x_n$ for all $n \in B$. This $p$ is in some $U_M$. pick $n \in B$ with $n > M$. Then $p \in U_M$ but $p = x_n \notin U_M$ contradiction. $\endgroup$ – Henno Brandsma Jun 21 '17 at 15:41
  • $\begingroup$ @MikhailKatz that is about converging sequences; it won't work in general $T_1$ spaces. $\endgroup$ – Henno Brandsma Jun 21 '17 at 17:29

Let $X$ be $T_1$. In the proof that "$X$ not countably compact implies $X$ not limit point compact" (the contrapositive) we start with a counterexample to countable compactness:

  • $\{U_n: n \in \mathbb{N}\}$ a countable open cover of $X$ without a finite subcover.

  • For each $n$, $\{U_1,\dots,U_n\}$ is not a cover of $X$, so pick $x_n \in X\setminus \cup_{i=1}^n U_i$. In particular:

$$(\ast) \forall n \ge m: x_n \notin U_m$$

Now define $A = \{x_n: n \in \mathbb{N}\}$ then $A$ is infinite. If not, there is some infinite set of indices and a point $p \in X$ such that $$\forall n \in B: x_n = p\text{,}$$

because some point $p \in A$ had to be chosen infinitely many times (pidgeon hole principle). But we have a cover and so $p \in U_m$ for some $m$, and then for $n \in B$ with $n > m$ (which surely exists as infinite subsets of $\mathbb{N}$ cannot lie completely below $m$), we would have simultaneously $p=x_n \in U_m$ and $x_n \notin U_m$ (by $(\ast)$), which is clearly absurd. So $A$ is indeed infinite.

Now, if $q \in X$ is any point of $X$, find some $m$ with $q \in U_m$. Then $U_m\cap A \subseteq \{x_1, \ldots x_{m-1}\}$, so is finite, by property $(\ast)$.

Call this finite set $F$. Then for each $f \in F$ such that $f \neq q$ pick a neighbourhood $U_f$ of $q$ such that $f \notin U_f$ by $T_1$-ness. Then $O:= U_m \cap \bigcap \{U_f: f \in F\setminus \{q\}\}$ is open (as a finite intersection of open sets, contains none of the $f \in F\setminus\{q\}$ so none of the $x_i \in (\{x_1, \ldots x_m\} \cap U_m)\setminus\{q\}$, and is is a subset of $U_m$ so contains none of the $x_n$ with $n > m$. So $O \cap A \subset \{q\}$ which means that $q$ is not a limit point of $A$. So $A$ can have no limit points at all, so $X$ is not limit point compact, as required.

We do need $T_1$ here for the equivalence, otherwise $X = \mathbb{N} \times \{0,1\}$ where $\{0,1\}$ has the indiscrete (trivial) topology and $\mathbb{N}$ the usual discrete one, is an example of a limit point compact space that is not countably compact.


In the proof that not countably compact implies not limit point compact: Here is a way to ensure that $A=\{x_n :n\in \mathbb N\}$ is not a finite set.

(i). Take $x_1\in X$ \ $U_1$ and let $f(1)=1.$ Now recursively:

(ii) .Let $F(n)=\max_{j\leq n}f(j).$

(iii). Take $x_{n+1}\in X$ \ $\cup_{j=1}^{F(n)}\;U_j$ and let $f(n+1)$ be the least (or any) $k$ such that $x_{n+1}\in U_k.$

If $m<n$ then $x_m\ne x_n$ because $x_n\not \in \cup_{j=1}^{F(n)}\;U_j\supset U_{f(m)}$ but $x_m\in U_{f(m)}$.

Observe that $f:\mathbb N\to \mathbb N$ is injective so $\lim_{n\to \infty}F(n)=\infty.$ So if $x\in U_m$ for some $m,$ then for all but finitely many $n$ we have $F(n)>m.$ And $F(n)>m$ implies $x_n \not \in U_m.$

  • 1
    $\begingroup$ It's always infinite by a pigeon hole argument, I thought of going recursive too (see edits), but realised it only complicated things. $\endgroup$ – Henno Brandsma Jun 22 '17 at 4:15

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