Compact Sets are Separable from Points in a Hausdorff Space

Last updated Nov 1, 2022

Let $X$ be a Hausdorff Topological Space. Then if $K \subset X$ is Compact and $y \not\in X$, then there exists $U, V \subset X$ Open so that

1. $K \subset U$
2. $y \in V$
3. $U \cap V = \emptyset$

Let $x \in K$. Because $X$ is Hausdorff, there exists Open $U_{x}, V_{x} \subset X$ so that $x \in U_{x}, y \in V_{x}$ and $U_{x} \cap V_{x} = \emptyset$. ${U_{x}}{x \in K}$ forms an Open Cover of $K$, so it can be reduced to a finite Open Subcover ${U{x_{i}}}{i=1}^{n}$ for some $n \in \mathbb{N}$. Take $$V = \bigcap\limits{i=1}^{n}V_{x_{i}} \text{ and } U = \bigcup\limits_{i=1}^{n} U_{x_{i}}.$$Then, $V$ is Open since it is just the finite Set Intersection of open sets. Because $y \in V_{x_{i}}$ for each $i \in [n]$, $y \in V$. $U$ is also Open since it is a Set Union of Open sets. Since ${U_{x_{i}}}{i=1}^{n}$ is an Open Cover of $K$, $K \subset U$. For each $i \in [n]$, Since $U{x_{i}} \cap V_{x_{i}} = \emptyset$, and $V \subset V_{x_{i}}$, so we have $U_{x_{i}} \cap V = \emptyset$. Then $V \cap \bigcup\limits_{i=1}^{n} U_{x_{i}} = V \cap U = \emptyset$ as desired. $\blacksquare$

• Mutually Disjoint
• Empty Set
• Natural Numbers