Question: Verify that if $E$ is a Banach Space, then the Vector subspace $V \in E$ is Banach (with restriction of the norm) if and only if it's closed.
I managed to prove that if, in the question conditions, $V$ is closed, then it's Banach.
But if $V$ is banach, i'm not able to prove that it is closed.
I tried to make the following work:
Suppose that $V$ isn't closed. So, there is a sequence $\{x_n\} \in V$ with ${x_n} \rightarrow x_0$ and $x_0 \notin V$. From that, i tried to prove that i can construct $x_0$ with elements of the sequence and get a contradiction, but i didn't manage to do it.
Also, i cannot fit the fact that $E$ and $V$ are Banach into solving the question. Maybe this info just doesn't matter (don't know).
So, if you have any hints, would be really helpful.
Thanks!
Best Answer
If $\{x_n\}\subset V$ converges to $z\in E$, then $\{x_n\}$ is a Cauchy sequence. Since $V$ is complete, there must be a limit $y\in V$. By the uniqueness of the limit, $y=z$ and so $z\in V$. As pointed out in the comments, this proof is valid in any complete metric space.