TITLE: Prove the set is closed with respect to its norm......
QUESTION [2 upvotes]: Let $V$ be a normed vector space over R.
Let $W$ be a proper closed subspace of $V$.
We say $w^*$ is a best approximation in $W$ to $v^* \in V$
if $\|v^*-w^*\| \leq \|v^*-w\|$ for all $w \in W$.
For a give $v^* \in V$, let $M$ be its set of best approximations in $W$.
Prove that $M$ is closed convex.

I proved that $M$ is convex.
But I'm having hard time proving that $M$ is closed with respect to the norm.
The following is my attempt to prove $M$ is closed with respect to its norm.

$$M = \{w ' \in W : \|v^*-w'\| \leq \|v^*-w\| \text{ for all }w \in W\}$$
Let {$w_n$} be a convergent sequence in $M$ converging to $w_o \in W$
So for all $\epsilon > 0 $, there exists $N$ such that for $n > N$,
we have $\|w_n - w_o\| < \epsilon$
Then we want to show that $w_o \in M$
That is, we want to show that $\|v^* - w_o\| \leq \|v^* - w\|$ for all $w \in W$.

I've been playing with triangular inequality.
But it doesn't work so far.
Can you help me with this?
Thank you.

REPLY [0 votes]: Fix $w\in W$.  We have for each n,  $\|v^*-w_n\|\le \|v^*-w\|$  Now, take the limit as $n\to \infty$, and use the fact that the norm is continuous to justify exchanging the limit with the norm, to conclude that $\|v^*-w_0\|\le \|v^*-w\|$
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