TITLE: Puzzle on the proof of that test function space is not metrizable
QUESTION [5 upvotes]: Rudin argues in "functional analysis", the test function space $D(\Omega)$ on nonempty open set $\Omega$ is not metrizable. Let $D_K$ be the subspace of $D(\Omega)$ consisting of functions with support contained in compact $K$. Rudin said that "it is obvious that each $D_K$ has empty interior relative to $D(\Omega)$". I don't understand after thinking for hours. Any help is appreciated.

REPLY [5 votes]: Assume that $X$ is a topological vector space, if $M$ is a vector subspace of $X$ with nonempty interior, then $M=X$.
So if $D_{K}$ has nonempty interior, then $D_{K}=D(\Omega)$, if $\Omega$ is not compact, then it is a contradiction.