Closed Hilbert subspace S with $S^\perp=\{0\}$

Question

Let $S \subseteq \mathcal{H}$ be a closed Hilbert subspace with $S^\perp=\{0\}$. Then determine if $S=\mathcal{H}$

I am trying to show this and intuitively it seems true because the only perpendicular vector is the zero-vector so it includes the span of all vectors of all basis.

However I am not sure how to show this mathematically and also - does it matter that S is closed or would the same be true?

Answer 1

Since $S$ is a closed subspace, we have $H=S \oplus S^{\perp}$. Since $S^\perp=\{0\}$, we derive $S=H.$

Answer 2

A basic theorem in Hilbert space theory says that if $S$ is a closed subspace of $H$ then any vector $x$ can be written as $y+z$ where $y \in S$ and $z \in S^{\perp}$. In this case $z=0$ so we get $x =y \in S$. Hence $S=H$.