Is the set $\{x_n=\cos (nt): n\in\mathbb{N}\}$ in $L_2[-\pi,\pi]$closed or compact? I don't know how to prove it.
The compactness of $\{x_n=cos nt\}_{n\in\mathbb{N}}\in L_2[-\pi,\pi]$
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sequences-and-series
functional-analysis
compactness
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0Probably you should also say what topology you mean. The sequence $x_n$ converges to zero in the weak topology. – 2012-12-10
2 Answers
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It's not compact, because it is a discrete subspace of $L^2$, and there are no countable discrete compact spaces. You could also say that if it were compact, there'd be a convergent subsequence $x_{\phi(n)}$, but $x_n$ tends weakly to zero, and $0$ is not one of the $x_n$.
However, it is closed.
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0Sorry, I misused the word "discrete". What I meant was that it has no limit points, because as in Davide's answer, no pair of points is closer than distance 2. – 2012-12-11
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As the elements are orthogonal, we have for $n\neq m$ that $\lVert x_n-x_m\rVert_{L^2}^2=2,$ proving that the set cannot be compact (it's not precompact, as the definition doesn't work for $\varepsilon=1/2$).
But it's a closed set, as it's the orthogonal of the even square-integrable functions.