Let $X$ be a normed space and $M$ be a linear subspace of $X$. Let $B_M$ be a closed unit ball in $M$ such that $B_M$ is closed in $X$. Then can we get $M$ is closed in $X$? Or other conclusions we can get from it.

Thank you in advance!

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    $\begingroup$ $M$ is the pre-image of $\Bbb K\times B_M$ under the continuous map $\Bbb K\times X\to X, \ (\lambda,v)\mapsto \lambda\cdot v$. So you find that if $B_M$ is closed in $X$ then $M$ is also closed in $X$. What you get in the end is that $B_M$ closed in $X$ $\iff$ $M$ closed in $X$. $\endgroup$ – s.harp Sep 5 '18 at 17:52
  • $\begingroup$ @s.harp Thanks for your comment. I am a little slow can you be more specific about why $B_M$ closed in $X$ then $M$ is closed in $X$? $\endgroup$ – Answer Lee Sep 5 '18 at 18:51
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    $\begingroup$ @AnswerLee Take any convergent sequence in $M$. "blow up" the unit ball (by multiplying it with a large scalar) so much that the convergent sequence lies entirely in the ball. The blown up ball is still closed and therefore contains the limit of the sequence. Since the ball is a subset of $M$, $M$ contains the limit of the sequence. $\endgroup$ – Michael Greinecker Sep 5 '18 at 19:59
  • $\begingroup$ @s.harp Why the pre-image of $\mathbb{K}\times B_M$ is $M$ and $\mathbb{K}\times B_M$ is closed? $\endgroup$ – Answer Lee Sep 5 '18 at 20:11
  • $\begingroup$ @s.harp I don't understand either. $\mathbb K\times B_M$ is in the domain of this map, so you can't take a pre-image of it. $\endgroup$ – Aweygan Sep 5 '18 at 20:56

Here's a proof using a nice norm identity: Suppose $(x_n)$ is a sequence in $M$ convergent to some $x\in X$. Without loss of generality, assume $x\neq0$, so we may assume $x_n\neq0$ for all $n$. By continuity of the norm, $\|x_n\|\to\|x\|$, so by the identity $$\left\|\frac{z}{\|z\|}-\frac{y}{\|y\|}\right\|=\frac{1}{\|z\|}\left\|(z-y)+(\|y\|-\|z\|)\frac{y}{\|y\|}\right\| \qquad (y,z\in X\setminus\{0\}),$$ it follows that $\frac{x_n}{\|x_n\|}\to\frac{x}{\|x\|}$. Since $B_M$ is closed, it follows that $\frac{x}{\|x\|}\in B_M$, whence $x\in M$ and therefore $M$ is closed.

  • $\begingroup$ Thanks for your answer. But I try to understand his answer first. Can you help me understand why the pre-image of $\mathbb{K}\times B_M$ is $M$ and $\mathbb{K}\times B_M$ is closed? $\endgroup$ – Answer Lee Sep 5 '18 at 20:42
  • $\begingroup$ I am sorry I am a little slow. But why without loss of generality, assume $x\neq0$, so we may assume $x_n\neq0$ for all $n$. $\endgroup$ – Answer Lee Sep 5 '18 at 21:05
  • $\begingroup$ For the first point, we're trying to show $x\in M$. But $0\in M$, so if $x=0$, we have nothing to do. For the second point, if $x\neq0$, then we can take some ball centered at $x$ which doesn't contain $0$. This ball contains all but finitely members of the sequence, so we can just cut off that first part of the sequence, and work with what we know is nonzero. $\endgroup$ – Aweygan Sep 5 '18 at 22:23

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