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

• Nov 19th 2012, 04:31 PM
machi
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 $Y$ of Banach space $X$ is complete if and only if $Y$ is closed in $X$"

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

from left to right "let $X$ is Banach space, $Y\subset X$. so, $Y$ is Banach space. consider of Banach space definition, every Cauchy sequence of $Y$ is converge to $x\in X$ then $Y$ is closed on $X$".
right to left "I am still totally confuse..."(Crying)
• Jan 8th 2015, 05:43 AM
Rebesques
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.