Let $u_n$ be a sequence in Hilbert space such that $\|u_n\|=1$ for all $n$, and $\langle u_n|u_m\rangle=0$ whenever $n\neq m$. Why is the following set closed: $\{0\}\cup \{u_n \mid n\geq 1\}$? Thanks for the help!
Help with proof of closedness of a set
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general-topology
hilbert-spaces
1 Answers
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Note that $\| u_n-u_m\|=\sqrt{2}$.
So for any two points $x$, $y$ of your set $S=\{0\}\cup\{u_n; n\ge 1\}$ you have $x\ne y \Rightarrow \|x-y\|\ge1.$
Hence every point of $S$ is an isolated point.
If $x_n$ is a convergent sequence of points from $S$, then $x_n$ must be eventually constant. (Otherwise it would not be Cauchy.) So it converges to some point of $S$.
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0Excellent. I was going to write more or less the same. I'll have to become faster! – 2012-10-15