Almost compact sets

Question

Update:

1. Q1 is answered in the comments.

2. I think that the usual arguments show that every relatively almost compact set in a space is closed in the space.

Original question:

A set $K$ in a space $X$ is almost compact if every open cover of $K$ has a finite subset $\cal F$ with $K\subseteq \bigcup_{U\in{\cal F}}cl({U})$.

Q1: Was this terminology used before, with the same or equivalent definition?

(I know the answer "Yes" for almost Lindel"of and almost Menger)

A set $K$ is relatively almost compact if the same holds, but only for open covers of the entire space $X$.

Recall that a set $K$ is relatively compact if every open cover of the entire space has a finite subcover of $K$. $K$ is relatively compact if and only if its closure in the space is compact.

Q2: Is there a similar characterization of relatively almost compact?

The closure of a relatively almost compact set is again relatively almost compact. Could we somehow get an absolute, nonrelative definition?

Answer 1

I don't specifically recall "almost compact" in the literature, but it's quite natural as it relates to "almost Menger" and "almost Lindelof".

In general, relative compactness is not equivalent to a set's closure being compact (sometimes called precompact). For example, in the particular point topology on a countably-infinite set, the singleton containing the particular point is relatively compact because it's compact. But its closure is the whole space, which is not compact. However, the claim does hold true for regular spaces (shown I believe first by Arhangelskii), so we probably should restrict our attention there.

Let $R$ be a relatively almost compact subset of a regular space $X$. We will show that $\overline{R}$ is compact (not just almost compact). Let $\mathcal U$ be a cover of $\overline{R}$. For each $x\in\overline{R}$ pick $x\in U_x\in\mathcal U$. Apply regularity to obtain $x\in V_x\subseteq \overline{V_x}\subseteq U_x$. Then we may cover $X$ using $\{X\setminus\overline{R}\}\cup\{V_x:x\in\overline{R}\}$. Apply relatively almost compact to obtain finite $F\subseteq \overline R$ with $R\subseteq\bigcup\{\overline{V_{x}}:x\in F\}$. Since $\bigcup\{\overline{V_{x}}:x\in F\}$ is closed, $\overline R\subseteq \bigcup\{\overline{V_{x}}:x\in F\}\subseteq\bigcup\{U_x:x\in F\}$. Since $\{U_x:x\in F\}$ is a finite subset of $\mathcal U$, we've shown $\overline R$ to be compact.

On the other hand, with or without regularity, $\overline R$ compact implies $R$ is relatively compact and thus relatively almost compact. So for regular spaces, relative almost compactness is equivalent to relative compactness and precompactness.

EDIT: One more result for regular spaces. Given almost compact $R$, cover $R$ with $\mathcal U$, where $x\in V_x\subseteq\overline{V_x}\subseteq U_x\in\mathcal U$ for each $x\in R$. Take $F\subseteq R$ with $\{\overline{V_x}:x\in F\}$ covering $R$, and therefore $\{U_x:x\in F\}$ covers $R$, showing $R$ is compact. Thus almost compactness is equivalent to compactness in regular spaces.