1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $x_n \not \in U_m.$

$\endgroup$
  • 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

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.