$X$ is c-closed iff every countably compact subset of $X$ is closed

Question

I need help to understand definition of c-closed. In here https://www.sciencedirect.com/science/article/pii/0166864180900279, author said that $X$ is c-closed iff every countably compact subset of $X$ is closed. Equivalently, every non-closed subset $A$ of $X$ contains a sequence which has no a cluster point in $A$.

I can not understand the equivalence. Is every non-closed containing a sequence which has no a cluster point in $X$? Then every space is c-closed?

I know it's wrong. For example, the ordinal space $\omega_1+1$ with the order topology is not c-space since $\omega_1$ is countably compact but isn't closed. How to understand this space isn't c-closed using second definition?

Answer 1

The cluster point of the non-closed set $A$ is considered to be in $A$, as you first wrote it.

The second definition is just the contrapositive of the first:
If every sequence of $A$ has a cluster point in $A$, then it is countably compact, hence it must be closed if $X$ is c-closed.

For the specific example, we would need to find a sequence in $\omega_1$ that tends to $\omega_1$ (the only missing point), which doesn't exist, as $\omega_1$ is regular.

Answer 2

A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.

The alternative formulation is

$(\ast$) For all $A \subseteq X$: if $A$ is not closed there exist $a_n \in A$ such that $A$ contains no cluster point for $(a_n)$.

The proposed equivalence is simply a matter of contrapositive reasoning:

Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($\ast$). For the reverse, suppose $X$ obeys the property $(\ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(\ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.

The example $X= \omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($\ast$) is:

There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.

And indeed $A = \omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($\ast$).

I think it's mostly a matter of taking the negation of the alternative definition in the right way.

Answer 3

Terminology: "Open" and "closed" mean "open in $X$" and "closed in $X$".

For one direction: Let $X$ be $c$-closed and let $A\subset X$ be non-closed. Then $A$ is not countably compact, so let $C=\{C_k: k\in \Bbb N\}$ be an open cover of $A$ with no finite sub-cover.

For $k\in \Bbb N$ let $D_k=\cup_{j\leq k}C_j$ and $E=\{D_k: \neg [\exists k'<k\;(A\cap D_{k'}=A\cap D_k)]\;\}.$

We can enumerate $E$ as $E=\{E_n:n\in \Bbb N\}$ where $A\cap E_n\subsetneqq A\cap E_{n+1}$ for each $n\in \Bbb N.$ For each $n\in \Bbb N$ choose $a_n\in A\cap (E_{n+1}\backslash E_n).$

Any $a\in A$ belongs to $E_{F(a)}$ for some $F(a)\in \Bbb N$ but the set $\{a_n:n>F(a)\}$ is disjoint from $A\cap E_{F(a)}.$ So $F(a)$ is a nbhd of $a$ such that $\{n: a_n\in E_{F(a)}\}$ is finite. So $a$ is not a cluster point of the sequence $(a_n)_{n\in \Bbb N}.$

Addendum: Re the space $\omega_1+1$ with the $\epsilon$-order topology: If $(a_n)_{n\in \Bbb N}$ is any sequence of members of $A=\omega_1$ then $\{a_n:n\in \Bbb N\} \subset B=(\cup \{a_n:n\in \Bbb N\})+1\in \omega_1.$ The sub$space $ $ B $ is compact & Hausdorff so $(a_n)_{n\in \Bbb N}$ has a cluster point $a\in B\subset A.$