# Under which $T$ axiom can you separate disjoint closed and compact set? Compact and compact set?

I am trying to figure out what is the lowest separation axiom needed that allows you to

1. separate disjoint closed and compact set via disjoint open sets, and

2. separate disjoint compact and compact sets via disjoint open sets.

Claim: Disjoint closed sets and compact sets in $T_3$ space and above can be separated via disjoint open sets

Pf. Take disjoint closed and compact sets $C,K$ respectively. Then since the space is $T_3$ (more importantly, regular), we can take disjoint open sets $V_k, U_k$ such that $k \in V_k$, where $k \in K$, and $C \subset U_k$

Then finite union of $V_k$ covers $K$, and finite intersection of $U_k$ covers $C$, those are disjoint open sets by construction.

What about compact sets? Is Hausdorfness enough?

Claim: Disjoint compact sets in $T_2$ space and above can be separated via disjoint open sets

Let $K_1,K_2$ be disjoint compact sets in Hausdorff space.

Then $\forall x \in K_1, y \in K_2$, there exists disjoint open sets $U_x, V_x$ such that $x \in U_x, y \in V_x$.

Then $\{U_x|x \in X\}, \{V_x| x \in X\}$ are open covers of $K_1, K_2$ with finite subcovers $\{U_{x_i}|x_i \in X, i \in {1, \ldots, n}\}, \{V_{x_i}|x_i \in X, i \in {1, \ldots, n}\}$.

We need to show that they are disjoint, let $z \in \cup U_{x_i} \cap \cup V_{x_i}$, then $z \in U_{x_i}, \forall x_i \in X$, and $V_{x_i}, \forall x_i \in X$. This is impossible since $U_{x_i}$ and $V_{x_i}$ are disjoint $\forall x_i \in X$

Are the above characterizations correct? Are there more generalizations?