Normal core of closed subgroup is closed

Statement

Suppose $G$ is a topological group and $H$ is a closed subgroup of $G$ . Then, the normal core of $H$ in $G$ (i.e., the largest normal subgroup of $G$ contained in $H$ , also defined as the intersection of all the conjugate subgroups of $H$ ) is also a closed subgroup of $G$ .

Given: A topological group $G$ , a closed subgroup $H$ of $G$ .

To prove: The normal core of $H$ in $G$ is also a closed subgroup of $G$ .

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	For any $g \in G$ , the conjugation map $x \mapsto g x g^{- 1}$ is a self-homeomorphism of $G$ .	Definition of topological group	$G$ is a topological group		PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
2	For any $g \in G$ , the conjugate subgroup $g H g^{- 1}$ is closed in $G$	self-homeomorphisms send closed subsets to closed subsets	$H$ is closed in $G$	Step (1)	Step-fact-given combination direct
3	The normal core of $H$ in $G$ , which is the intersection $⋂_{g \in G} g H g^{- 1}$ , is closed in $G$ .	arbitrary intersection of closed subsets is closed		Step (2)	Step-fact combination direct.