Hilbert space question; show Y is complete iff closed

[mathplease]

I would like a second opinion on my answer to this question as I'm confusing myself thinking about my proof. Any input is appreciated

Homework Statement

"Let (X, ||.||) be a complete normed linear space and Y $$\subset$$X be a non-empty subspace of X. Then (Y, ||.||) is a normed linear space. Show that Y is complete if and only if it is closed."

Homework Equations

convergent sequence: http://mathworld.wolfram.com/ConvergentSequence.html"

cauchy sequence: http://mathworld.wolfram.com/CauchySequence.html"

complete: a normed linear space in which every cauchy seq is convergent is complete

closed: (X,||.||) is a normed linear space. A is closed if {x_n} $$\subseteq$$ A $$\subseteq$$ X and x_n-> x then x $$\in$$A.

The Attempt at a Solution

Let {y_n} be a Cauchy sequence in Y. Since (X,||.||) is complete, y_n converges to y$$\in$$X. Assuming Y is closed: y$$\in$$Y.
Hence, Y is complete.

Conversely,
assume Y is complete. Let {y_n} be a convergent sequence in Y. Since convergent sequences are Cauchy, {y_n} is a Cauchy sequence.
Since Y is a complete normed linear space y_n$$\rightarrow$$y $$\in$$Y (Cauchy sequences converge).
Hence Y is closed.

Therefore Y is complete if and only if it is closed.

 

[jbsteele]

Second Opinion: Your proof looks correct. The main idea is that if a Cauchy sequence converges, then it must converge to a point in Y since Y is closed. And if Y is complete, then a sequence in Y will always converge. So the two concepts are related, and your proof captures that.