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

I was looking for an example of a bounded and closed set which is not compact. Considering $l^2$ and looking to a set $K$ of canonical basis $e_i=(0,...,1_i,...,0)$. This is bounded, Is it true that under the norm function pre-image of {$1$} is $K$? This will prove that it is closed. ||$x$||=$\sqrt{| x_i|^2}$

share|cite|improve this question
You can observe that $\lVert e_i - e_j \rVert = \sqrt 2$ whenever $i \neq j$. This shows that a convergent sequence must be eventually constant equal to $e_i$ for some $i$, hence your set $K$ is closed. – Martin Mar 2 '13 at 8:17
up vote 2 down vote accepted

No, there are many vectors with norm one that are not one of the canonical base: there are $-1$ coordinates to consider as well, and take any point on the circle and extend with $0$'s, a point on the sphere (in $\mathbb{R}^3$) and extend by $0$'s etc.

But the set of all vectors of norm 1 in in $\ell^2$ is a good example of a closed (by your inverse image of norm idea) and bounded (by definition) set that is not compact, and you can use the unit vectors for that...

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.