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

#### Fernando Revilla

##### Well-known member
MHB Math Helper
I quote an unsolved question posted in MHF (November 19th, 2012) by user machi.

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..."

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
Let $$X$$ be a Banach space and $$Y$$ a subspace. We have to prove that $$Y$$ complete $$\Leftrightarrow\;Y$$ closed.

$$\Rightarrow)$$ If $$x\in \mbox{cl}(Y)$$ there exists a sequence $$x_n$$ in $$Y$$ which converges to $$x$$ and as a consquence is a Cauchy sequence (in $$X$$ and $$Y$$). But $$Y$$ is complete, so $$x_n$$ converges in $$Y$$ to a point $$x'$$. According to the uniqueness of the limit, $$x=x'\in Y$$. That is, $$\mbox{cl}(Y)\subset Y$$ and hence, $$Y$$ is closed.

$$\Leftarrow)$$ Let $$x_n$$ be a Cauchy sequence in $$Y$$. The limit $$x$$ of $$x_n$$ exists in $$X$$ because $$X$$ is complete. As $$Y$$ is complete, contains all its limit points, so $$x\in Y$$ and so, $$Y$$ is complete.

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#### amin

##### New member
in the second part of the solution i get confused by " as Y is complete " here in the second part we want to prove it is complete.
so it should be Y is closed then it contain all of its limit points
if Y\subset X, Y closed \iff it contains all of its limit points its a theorem

#### Opalg

##### MHB Oldtimer
Staff member
in the second part of the solution i get confused by " as Y is complete " here in the second part we want to prove it is complete.
so it should be Y is closed then it contain all of its limit points
if Y\subset X, Y closed \iff it contains all of its limit points its a theorem
That looks like a simple misprint. It should say "as Y is closed". The proof then works smoothly.