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

Riesz' lemma gives us that in infinite-dimensional spaces no ball is compact, but what about the sphere $\{x \in X : \|x\| = 1\}$? Can we say something about the compactness of the sphere in infinite-dimensional spaces?

(I guess the sphere is also not compact and I think one can also show this by constructing a sequence with Riesz lemma that has no convergent subsequence). Is this idea correct?

share|cite|improve this question
Riesz's Lemma says given a proper closed linear subspace $Y$ of the normed linear space $X$ and $0<\theta<1$, there is an element $x$ of norm $1$ so that $\Vert x-y\Vert>\theta$ for all $y\in Y$. One then can construct a sequence in the unit sphere of $X$ that is separated. – David Mitra Jan 2 '14 at 14:31
up vote 10 down vote accepted

Suppose that the unit sphere $S_X$ of $(X, \| . \|_X)$ is compact. Then the unit ball of $X$ is the image of the compact set $[0,1] \times S_X$ by the continuous map $(t, v) \mapsto tv$, and hence is compact.

share|cite|improve this answer
particularly nice proof, thanks. – user66906 Jan 2 '14 at 15:23

Hint: In a metric space, sequential compactness and compactness are equivalent. Now consider a sequence consisting of unit length basis vectors.

share|cite|improve this answer

Sounds right to me. The sequence of points $(1,0,0,\ldots), (0, 1, 0, \ldots), (0, 0, 1, \ldots), \ldots$ is a sequence of points on the sphere that has no convergent subsequence, because the distance between any two of the points is $\sqrt{2}$.

share|cite|improve this answer

Your Answer


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