I was hoping someone could review my proof for correctness. Thanks in advance!

Let $X$ be a $T_1$ space. Let $X$ be limit point compact. Suppose $X$ is not countably compact.

Then take the countable open covering without a finite subcover : $\mathcal{U} = \{ U_n \mid n \in Z_+ \}$

Then since no finite subcollection of $\mathcal{U}$ can cover $X$ we can define a sequence $(x_n)$, that obeys:

$$\forall n \in Z_+: x_{n+1} \notin U_1, U_2, \ldots U_n$$

Again, each $x_n$ exists since if we were unable to choose such an $x_n$ then we have reached a finite subcover $U_1, \ldots , U_n$. And by hypothesis, no finite subcover can be found for our open covering $\mathcal{U}$

Now $B = \{x_n: n \in Z_+\}$ is an infinite set. Therefore $B$ has a limit point, call it $z$.

Since $\mathcal{U}$ is an open covering, some $U_n$ must contain $z$.

(1) Since $X$ is a $T_1$ space and $z$ is a limit point of $B$, $U_n$ must intersect $B$ in an infinite number of points.

But for any $m > n$ we have $x_m$ isn't a member of $U_n$ by construction of the sequence $(x_n)$. So then we are left with $U_n$ only able to intersect B in a finite number of points -- namely at most n points. A contradiction since $X$ is a $T_1$ space and $U_n$ must intersect $B$ in infinitely many points.

Therefore, $X$ must be countably compact.

  • $\begingroup$ Can you explain where exactly did we get a contradiction for $T_1$ space? $\endgroup$ – Aniruddha Deshmukh Mar 12 at 4:09
  • $\begingroup$ see (1). This is a property of T1 spaces. A neighborhood of a limit point of a set A must intersect A in infinitely many points. $\endgroup$ – H_1317 Mar 12 at 4:16
  • $\begingroup$ Actually in any topological space, a neighbourhood of a limit point must intersect the set $A$ at infinitely many points. $\endgroup$ – Aniruddha Deshmukh Mar 12 at 4:36
  • 1
    $\begingroup$ No that is not true in general. Take the set X = {1,2,3} with the topology = { {1,2,3}, {1,2}, {1,3}, {1}, empty set}. We then have that 2 is a limit point of the set {1}. Since every neighborhood containing 2 also intersects the set {1}. Clearly 2 only intersects the set {1} in finitely many points @AniruddhaDeshmukh $\endgroup$ – H_1317 Mar 12 at 6:03

This proof is fine (after the formatting I did); it's a bit easier to assume that $U_n \subseteq U_{n+1}$ for all $n$, which can easily be achieved WLOG, using the cover $V_n = \bigcup_{i=1}^n U_i$ instead: if the $U_n$ have no finite subcover, the same holds for the increasing open cover $V_n$; this also guarantees that $x_n \neq x_m$ for $n \neq m$, amking $B$ infinite trivially.

Note that this proof shows that in any space $X$ (not necessarily $T_1$)

$X$ is countably compact iff every (countably) infinite subset $A$ of $X$ has an $\omega$-limit point.

where $x$ is a strong limit point of $A$ when every open neighbourhood of $x$ contains infinitely many (hence the $\omega$) points of $A$. The proof of the forward direction is not too hard, but nice, see the proof I gave here, e.g.

The latter characterisation via limit points can also be done for "plain" compactness:

$X$ is compact iff every infinite subset $A$ of $X$ has a point $p$ such that for every neighbourhood $O$ of $p$: $|O \cap A| = |A|$ (where the bars denote cardinality); such a $p$ is called a point of total condensation for $A$, it's a special kind of limit point.

(it's a somewhat old-fashioned characterisation that Tychonoff used to show his theorem on the product of compact spaces being compact).

  • $\begingroup$ This post is interesting and thank you for reviewing my proof. I do have some questions: 1.) how can we assume each U_n is a subset of U_n+1 ? This seems like a very specific assertion for an arbitrary countable open covering with no finite subcovering. 2.) it sort of seems like your first yellow block using ur strong limit point definition is essentially just taking the useful property from a T1 space and renaming it to a “strong limit point”. Or am i misreading ? $\endgroup$ – H_1317 Mar 12 at 16:43
  • $\begingroup$ @H_1317 Like I said: if we have a countable counterexample cover (without a fnite subcover) we also have an increasing one. So we might as well work with that one if it's more convenient (all we need is just one contradiction)... $\endgroup$ – Henno Brandsma Mar 12 at 22:51
  • $\begingroup$ @H_1317 The notion of strong limit point is independently motivated. But indeed in a $T_1$ space it's equivalent to being a limit point. It "explains" why in a $T_1$ space we get away with limit point compactness being equivalent to countable compactness, while in general spaces, to prove a similar equivalence, we need its full "force", being a strong (or $\omega$-) limit point. $\endgroup$ – Henno Brandsma Mar 12 at 22:54
  • $\begingroup$ Why is it that a countable covering without a finite subcover must have an increasing cover for any space? @HennoBrandsma $\endgroup$ – H_1317 Mar 12 at 22:59
  • 1
    $\begingroup$ @H_1317 If $\{V_1, V_2, V_3,\ldots, \}$ is a countable open cover of $X$ without a finite subcover, define $U_n = \bigcup_{i=1}^n V_i$ for all $n$. Then $\{U_1,U_2, U_3, \ldots\}$ is a countable open cover of $X$ without a finite subcover that moreover obeys $U_n \subseteq U_{n+1}$ for all $n$. $\endgroup$ – Henno Brandsma Mar 12 at 23:02

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.