finite dimention subspaces of infinite dimention hilbert space are closed?

Question

Let H is a hilbert space of infinite dimention, and $ V \subset H $ of finite dimention, can we show that V is closed?

I know if H is of finite dimention, then V is closed, but what if H is infinite dimention?

Do it first for $\dim V=1$, and then proceed by induction on the dimension. — Siminore, Jun 25 '14 at 12:55
Any finite dimensional subspace of a hilbert space is closed. — AnyAD, Jun 25 '14 at 13:03

score 1 · Accepted Answer · answered Jun 25 '14 at 13:20

Here a topological proof:

Let $V$ be the finite dimensional subspace of $H$. We prove that its complement is open.

Let $x\notin V$ and let $R=2||x||$.

Let $D_V(0,R)$ and $B_V(0,R)$ denote the closed and open balls of radius $R$ in $V$ (and $D_H(0,R)$ and $B_H(0,R)$ those in $H$).

$V\cap D_H(0,R)=D_V(0,R)$ is compact because closed bals are compact for finite dimensional spaces.

Since $H$ is Hausdorff, then $D_V(0,R)$ is closed in $D_H(0,R)\subset H$.

Since $x\notin V$ there is an open neigborhood $U$ of $x$ disjoint from $D_V(0,R)$. Now, $U\cap B_H(0,R)$ is an open neighborhood of $x$ disjoint from $V$.

finite dimention subspaces of infinite dimention hilbert space are closed?

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