Are all compact sets in a separable locally compact space G_delta sets?

Question

In Halmos' Measure theory Theorem E from $\S$50 states that a compact set $C$ in a separable locally compact space $X$ is a $G_\delta$ set. The proof goes by the following logic: for $x \in X \setminus C$ let $U(x)$ and $V(x)$ denote some disjoint open subsets of $X$ such that $C \subset U(x)$ and $x \in V(x)$. It is clear that $C = \bigcap_{x \in X} U(x)$. Thus, if we manage to take a countable subcover from the cover $\{V(x)\}_{x \in X}$ of $X \setminus C$, then it follows that $C$ is $G_\delta$. Halmos states that such subcover exists due to separability of $X$.

The question is, how can one deduce that separability implies the existence of such subcover? As far as I can understand, it is not possible to conclude that $X$ is Lindelof if it is locally compact and separable (without metrizability or something else). If this is the case, then the follow-up question is: is the statement of the theorem correct? Maybe Halmos meant to prove this fact under the assumption of $X$ having countable base.

Answer 1

As stated above this is wrong: $\beta \mathbb N$, the Stone-Cech compactification of the integers, is compact and separable. Let $x \in \beta \mathbb N \setminus \mathbb N$. Then $\{x\}$ is compact, but not $G_\delta$, since in a compact space $G_\delta$-points have countable neighborhood bases (see Engelking, General topology, 3.1.F and 3.6.17).

Note that this example also shows that Lindelöf does not imply this statement. It holds, however, for hereditarily Lindelöf T2 spaces, since then each closed subset is $G_\delta$.

Answer 2

Another example, $X = [0,1]^A$ with $|A| = |\mathbb R|$. Then $X$ is separable; a singleton in $X$ is closed but not $G_\delta$.

Answer 3

This is the ground-breaking text:

Halmos, Paul R., Measure theory, (University Series in Higher Mathematics) New York: D. Van Nostrand Co., Inc.; London: Macmillan & Co., Ltd., XII, 304 p. (1950). ZBL0040.16802.

Note the date: 1950.

From this book:
§0. Prerequisites
page 3

A space $X$ is separable if it has a countable base.

So, of course, Theorem E of §50 is correct.

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Remark. This was before Halmos invented the word "iff" for use in definitions, so he still uses "if", and the reader must understand that (because it is a definition) it means "if and only if".

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Some history.
In 1907, in his thesis, when Fréchet introduced the term separable, it was for the (not-yet named) "metric space" setting. In subsequent years, mathematicians found that (in metric space) there are several equivalent definitions for "separable".

A generation later, "topological space" was developed. These definitions were not equivalent in that setting. Mathematicians continued to use the term "separable" for one of these, but at first they did not always choose the same one. (So of course they included definitions in their papers and books, such as Halmos did.) In due course, we have agreed on the terminology.

[When I was in graduate school, in the 1970s, one of the older professors (George Mackey) used "separable" to mean "the topology has a countable base". By then, we students knew the "modern" terminology, so it was important to know what Mackey meant.]