A subspace of Banach space is complete if and only if it is closed.

I have a problem to proof this theorem, Anyone can help for detail.

"**A subspace $\displaystyle Y$ of Banach space $\displaystyle X$ is complete if and only if $\displaystyle Y$ is closed in $\displaystyle X$"**

I have an idea to prove this theorem, but I am not sure about this and I can't wrote it for detail.

Please correct my answer,

from left to right "let $\displaystyle X$ is Banach space, $\displaystyle Y\subset X$. so, $\displaystyle Y$ is Banach space. consider of Banach space definition, every Cauchy sequence of $\displaystyle Y$ is converge to $\displaystyle x\in X$ then $\displaystyle Y$ is closed on $\displaystyle X$".

right to left "I am still totally confuse..."(Crying)

Re: A subspace of Banach space is complete if and only if it is closed.

For the forward implication, suppose Y is complete. You want to show it is closed.

Suppose that a sequence in Y converges in X. We will show that its limit belongs to Y.

This is somewhat obvious, since a convergent sequence is a Cauchy sequence, and all Cauchy sequences in Y converge in Y.

For the inverse implication, suppose Y is closed in X. We show that Y is complete.

Consider a Cauchy sequence in Y. Since this also belongs to X and X is a Banach space, the sequence converges.

But Y is closed in X, and so the limit of the sequence must belong to Y. This means Y is complete.

Note that the assumption that Y is a subspace is not necessary.