Relative compactness implies relative countable compactness?

Question

By using the fact that compactness implies countable compactness, I think that relative compactness implies relative countable compactness in any topological space. Am I right?

Thank you so much!

Answer 1

Yes, if your definitions are as mine:

• relatively compact : the closure is compact,
• relatively countably compact: the closure is countably compact,
• countably compact: every countable cover has a finite subcover,

The implication is trivial.

[added after comment:] If you define relatively countably compact as: every infinite set has an accumulation point in the surrounding space, the proof is a bit more involved:

Suppose $A$ is relatively compact in $X$, then $\overline{A}$ is compact. Take any infinite subset $B$ of $A$. Then as $B \subset \overline{A}$ as well, and in a compact space every infinite set has an accumulation point (I suppose you know this?), there is some $x \in \overline{A}$ that is an accumulation point of $B$, and this is the required point to show that $A$ is indeed relatively countably compact in your definition.

We use: If $X$ is compact and $B \subset X$ is infinite, $B$ has an accumulation point in $X$. Otherwise, every $x \in X$ has a neighbourhood $U_x$ such that $U_x \cap B \subset \{x\}$. This defines a cover of $X$, which has a finite subcover $U_{x_1},\ldots,U_{x_n}$, but then $$B = B \cap X = B \cap (\bigcup_{i=1}^n U_{x_i}) = \bigcup_{i=1}^n (B \cap U_{x_i}) \subset \{x_1,\ldots.x_n\} $$ which contradicts $B$ being infinite. (Just for completeness.)