Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Probably you should also say what topology you mean. The sequence $x_n$ converges to zero in the weak topology. – GEdgar Dec 10 '12 at 22:33
up vote 4 down vote accepted

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.

share|cite|improve this answer
The easiest way to see it is closed, is to remember the fact that the $\cos(nt)$ and $\sin(nt)$ together (and properly renormalised) form a Hilbert basis for $L^2$. From there on it is rather straightforward. – Olivier Bégassat Dec 10 '12 at 21:23
Easier still: since it is discrete, it has no limit points, so vacuously it contains all its limit points, hence is closed. – Nate Eldredge Dec 10 '12 at 22:26
@NateEldredge I don't understand your comment, are you saying it is a closed subspace because it is discrete? That's not true, so what do you mean? – Olivier Bégassat Dec 11 '12 at 0:52
Sorry, 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. – Nate Eldredge Dec 11 '12 at 1:20

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.