Divergence on definitions of limit point compactness

Question

A topological space $\Omega$ is said to be weakly sequentially compact if every sequence in $\Omega$ has a limit point.

---

A topological space $\Omega$ is said to be limit point compact if every infinite subset of $\Omega$ has a limit point.

Are these two definitions equivalent? I'm doing a summer course on general topology and my professor just started talking about compactness, and the first definition is the one he uses, the second is from Munkres' textbook. My professor said they were the same thing but I'm not completely convinced yet and it's much more pleasant to work with the second one. It's clear the first definition implies the second one, but I'm not sure about the converse.

I think there's a problem in saying the definitions are equivalent because the second one deals with infinite subsets and the first only cares about sequences, so I could have a sequence that's eventually constant or only has a finite number of terms and it would have a limit point, which I think isn't necessarily true if all we're given is the second definition.

Am I right or are they really equivalent?

Answer 1

They are not in general equivalent: let $X=\mathbb{N} \times \{0,1\}$ in the product topology were the natural numbers carry the discrete topology and $\{0,1\}$ the trivial/indiscrete one.

If $A \subseteq X$ is infinite and $(n,i) \in A$, then $(n,i')$ (where $i' \neq i$) is a limit point of $A$, as every open set containing $(n,i')$ also contains $(n,i)$, so $O \setminus \{(n,i')\}$ intersects $A$.

If $(x_n)$ is a bijective enumeration of $X$ (say $x_1 = (1,0), x_2=(1,1), x_3=(2,0), x_4 = (2,1), \ldots$ it's clear that every point $(n,i)$ in $X$ has a neighbourhood $\{n\} \times \{0,1\}$ that only contains two points of the sequence, so the sequence has no limit point in $X$.

In any space $X$ that is weakly sequentially compact we can see that every infinite subset $A$ of $X$ has a point $p$ such that every neighbourhood of $p$ intersects $A$ in an infinite set. ($p$ is then called an $\omega$-limit point of $A$), so quite a bit stronger than just limit point compact.

If $X$ is $T_1$ (so finite sets are closed) then a limit point compact $X$ has the property that every infinite subset has an $\omega$-limit point in $X$. (thi is sometimes called "strongly limit point compact" or $\omega$-limit compact or something similar)

And so for $T_1$ spaces the notions are equivalent, but not quite in general.

For all spaces strong limit point compactness is equivalent to "every countable open cover of $X$ has a finite subcover" (something that my example space indeed fails), which is usually called "countably compact" (i.e. compactness for countable covers).

Answer 2

The two are equivalent (edit: in metric spaces or first countable spaces). Here's one direction, briefly: if $\Omega$ is limit-point compact, then consider a sequence $(x_n)$. The range $S=\{x_n: n\in \mathbb{N}\}$, if it is finite, trivially, has a limit point. If it is infinite, then since $\Omega$ is limit-point compact, we can find a limit point $x$ of $S$ that is in $\Omega$.

To emphasize that limit points are either elements of the space or the underlying ambient space: if $\Omega =[0,1]$ and $x_n=1/n$ then $0$ is a limit point of $S=\{x_n : n\in \mathbb{N}\} \subset \Omega\}$ but $0$ is not an element of $S$, while $0\in \Omega$. Compare to if $\Omega=(0,1]$ is considered as a subspace of $\mathbb{R}$ then $0$ is a limit point of $S$ still but now $0$ is in neither $S$ nor $\Omega$, we just know $0\in \mathbb{R}$. Exactly which sets/spaces limit points are elements of is inextricably tied up with the notions of closedness, completeness. etc.