The union of a set with its complementary components that have compact closure

Question

Let $X$ be a topological space and $A$ be a subset of $X$. Let $A'$ be a subset of $X$ such that $A'$ contains $A$ and all those subsets of $X\setminus A$ which are maximal connected sets and whose closures are compact.

Which one is correct?

1. $A'$ is compact
2. $A'$ is connected.
3. $A''=A'$
4. $A'=X$.

If I take $X$ to be an infinite discrete space then first two options get eliminated but I don't know what to do with the last two.

Answer 1

4 is false. Take $X = \mathbb{R} \backslash \{-1,0,1\}$ and $A = (0,1)$. Then, $A' = (-1,0) \cup (0,1) \neq X$. The crucial point is that connected components need not have compact closure.

3 is true. It's clear that $A \subseteq A' \subset A''$. Show that the connected components of $X\backslash A'$ are connected components in $X \backslash A$. Since they aren't subsets of $A'$, they do not have compact closures. Hence, $A''$ does not contain them either.