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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
Please correct my answer,

from left to right "let [tex]X [/tex]is Banach space, [tex]Y\subset X[/tex]. so, [tex]Y[/tex] is Banach space. consider of Banach space definition, every Cauchy sequence of [tex]Y[/tex] is converge to [tex]x\in X[/tex] then [tex]Y[/tex] is closed on [tex]X[/tex]".
right to left "I am still totally confuse..."
 
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Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Let [tex]X[/tex] be a Banach space and [tex]Y[/tex] a subspace. We have to prove that [tex]Y[/tex] complete [tex]\Leftrightarrow\;Y[/tex] closed.

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

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

New member
Oct 4, 2017
1
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
Feb 7, 2012
2,703
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.