Compactness: Opens vs. Neighborhoods

Question

Disclaimer: This thread is a record of thoughts.

Discussion Given a compact set.

Do mere neighborhood covers admit finite subcovers? $$C\subseteq\bigcup_{i\in I}N_i\implies C\subseteq N_1\cup\ldots N_n$$ (The idea is that neighborhoods are in some sense fat.)

Application

Given a locally compact space.

Every compact set has a compact neighborhood base: $$C\subseteq U:\quad N\subseteq U\quad(C\subseteq N^°)$$ (The above would give clues how to prove this.)

Answer 1

Counterexample

Consider the compact set: $[0,1]\cup[2,3]$

Cover it by neighborhoods: $$(x-\varepsilon,x+\varepsilon)\cup\{x+2\}\quad(0\leq x\leq1)$$

Then it admits no finite subcover!

Discrepancy

Consider the compact set: $\{1\}\cup\{-1\}$

Cover it by neighborhoods for all points: $$(1-\varepsilon,1+\varepsilon)\cup(-1-\varepsilon,-1+\varepsilon)\quad(1-\varepsilon,1+\varepsilon)\cup\{-1\}$$

Then it admits a bad finite subcover!

Essence: The power of opens lied in them being neighborhoods of its point.

Answer 2

Call a space $X$ really, truly compact if every cover of $X$ by sets with nonempty interior has a finite subcover.

It is not hard too show that $X$ is really, truly compact if and only if $X \setminus U$ is finite for every nonempty open $U \subseteq X$ (i.e., the topology on $X$ is coarser than the co-finite topology on $X$):

If $X$ has a nonempty open $U \subseteq X$ such that $X \setminus U$ is infinite, then $\{ U \cup \{ x \} : x \in X \setminus U \}$ is a cover of $X$ by sets with nonempty interior with no finite subcover.

If $X \setminus U$ is finite for every nonempty open $U \subseteq X$, consider any cover $\mathcal A$ of $X$ by subsets of $X$ with nonempty interior. Pick any $A_0 \in \mathcal A$. Then $X \setminus A_0 \subseteq X \setminus \operatorname{Int} (A_0)$ is finite, say $\{ x_1 , \ldots , x_n \}$, so we may pick $A_1 , \ldots , A_n$ from $\mathcal A$ so that $x_i \in A_i$. Then $\{ A_0 , A_1 , \ldots , A_n \}$ is a finite subcover of $\mathcal A$.

From this it follows that a T₁-space is really, truly compact if and only if it has the co-finite topology, and therefore the really, truly compact subsets of $\mathbb R$ are exactly the finite subsets.