Show that if a vector subspace of a Hilbert space is closed, then it is a Hilbert subspace.
Here is what I have so far. Any comments or hints are greatly appreciated.
Let $H$ be a Hilbert space and $S$ be the vector subspace of $H$. Since $H$ is closed that means $H$ contains all of its limit points. Let $(u_n)$ be a Cauchy sequence in $S$. Then $(u_n)\to U\in H$ (since $S$ is closed) where $U$ is a limit point of $S$. Thus $S$ is a Hilbert subspace.
Best Answer
Remember, just because a sequence is Cauchy in a metric space, doesn't necessarily mean that it converges, so I don't think you want to rely on the fact that S contains it's limit points for the proof, rather you probably want to use the fact that H is a Hilbert space and thus complete, meaning all cauchy sequences in H end up converging in H. From this perspective the proof isn't too difficult. As H is a Hilbert space, it's a complete metric space, furthermore, as S is a closed subspace of a complete metric space, it must then also be complete in regards to it's metric (whether or not you've shown this in the past, I'm not sure, but the proof itself isn't too difficult), thus making it a Hilbert space.