I know the following version of Borel density theorem:
If $G$ is a connected real Lie group such that every continuous homomorphism from $G$ to a compact group is trivial, and if $H$ is a closed subgroup such that $G/H$ admits a finite $G$-invariant measure, and if $L$ is a closed subgroup of $G$ which contains $H$ and has a finite number of connected components, then $L=G$.
I am aware of the following version:
If $G$ is a connected semisimple real algebraic group, and $H$ is a subgroup with the same properties as in the first version, then $H$ is Zariski dense in $H$.
Does the second version follow from the first? What facts about real algebraic groups and their connections to Lie groups do I need to know to understand this implications? (I know more about Lie groups than I know about algebraic groups).
The Zariski adherence of $H$ is a closed subgroup (which contains $H$) with finitely many connected components. You can apply the last part of the first version to conclude that the Zariski adherence of $H$ is $G$.