Take the tour ×  I'm trying to prove that $X$ is closed in
$X''$, where $X$ is a Banach space. I know that $X$ is
embeddable in $X''$.  If the isomorphism was bijective, I could
show that $X$ is closed since $X''$ closed and closeness is
preserved. But in general, it isn't bijective.  Which approach
should I take? Hints only please, if possible.  share|improve
this question add comment  1 Answer  up vote 2 down vote
accepted  Sketch: $X$ is complete. The embedding $(Jx)(\varphi)
= \varphi(x)$ is an isometry, so $J(X)$ is complete. Complete
subspaces are always closed.  share|improve this answer In fact,
a complete subspace of any (metric) space is closed. –  Chris
Eagle Dec 5 '12 at 22:48 @ChrisEagle thanks. Corrected. –  kahen
Dec 5 '12 at 22:49 add comment  Your Answer