9
$\begingroup$

It is known that in the Banach space with infinite dimension, the closed unit ball is not compact. I am just wondering if it is possible for the surface of the ball, say $\|x\|=1$ can be compact.

$\endgroup$
  • 5
    $\begingroup$ For the record, the usual proof of noncompactness of unit balls exhibits an infinite family of unit vectors which are pairwise separated, so the same proof actually shows the unit sphere is not compact. $\endgroup$ – Wojowu Nov 8 '17 at 20:10
  • $\begingroup$ Is "surface of the ball" just a complicated way to say "sphere"? $\endgroup$ – Pedro A Nov 8 '17 at 22:05
22
$\begingroup$

No, because then the closed unit ball would be compact, since it is the image of the map$$\begin{array}{ccc}[0,1]\times\{x\,|\,\|x\|=1\}&\longrightarrow&\text{your space}\\(t,x)&\mapsto&tx.\end{array}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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