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Let $A$ be a normed space and $B$ a finite dimensional subspace of $A$. Let $a\in A$, then we define the distance of $a$ to $B$ as $d(a,B)=\inf_{b\in B}||a-b||$.

How do I prove that there exists a $b_a\in B$ such that $d(a,B)=||a-b_a||$?

I have no idea where to begin this proof, so I would greatly appreciate a hint to start me off.

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If $d(a,B) = \eta$, there exists a sequence such $(b_n)_n \in B^{\mathbb{N}}$ such that $||b_n - a|| \rightarrow \eta$. Therefore the sequence is a bounded sequence of a finite dimensional normed space, hence Bolzano-Weirstrass yields that there exists a subsequence $(b_{n_k})_k$ that converges to $b_l \in B$, which finishes the proof.

Hermès
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  • Do you mean $b_a$ instead of $b_l$? –  Oct 02 '16 at 12:57
  • @BananachHammock yes indeed ;). Well actually $b_l = b_a$, which is not exactly the same statement. – Hermès Oct 02 '16 at 12:58
  • Thank you for your help, but I'm not sure why the existence of a subsequence $(b_{n_k})_k$ finishes the proof –  Oct 02 '16 at 13:00
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    @BananachHammock because $||b_a-a||\leq ||b_a-b_{n_k}||+||b_{n_k}-a||\rightarrow 0+\eta=\eta$ as $k\rightarrow \infty$ – Peter Melech Oct 02 '16 at 13:17