# showing subset of $l^2$ closed

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. ||$x$||=$\sqrt{| x_i|^2}$

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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

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...