7
$\begingroup$

More specifically, let $Y= X \cup \{\infty\}$ and declare open sets to be the usual open sets in $X$ together with those that are of the form $(X-C)\cup \{\infty\}$, where $C$ is norm closed and norm bounded. So $Y$ is formed in the same way that the one point compactification is when $X$ is finite dimensional.

Clearly if $X = \Bbb R^n$ we have $Y \cong S^n$, which motivates the question. Note that all separable Banach spaces are homeomorphic, so I think $Y$ should not depend on which Banach space is chosen (which makes the Hilbert space the natural candidate to try to work with.)

Any references to a paper/text with a proof or a sketch of a proof (or disproof) are greatly appreciated.

Thanks in advance!

$\endgroup$
  • $\begingroup$ Not that it answers the question, but X is already homeomorphic to its unit sphere without an added point; see here. $\endgroup$ – Michael Greinecker Jun 9 '18 at 23:02
  • $\begingroup$ @MichaelGreinecker Thank you for the link. That is interesting (I should have guessed that would be true since the sphere is contractible.) $\endgroup$ – 3-in-441 Jun 9 '18 at 23:46
  • 1
    $\begingroup$ Why wouldn't the stereographic projection work? $\endgroup$ – Berci Jun 10 '18 at 0:15
  • $\begingroup$ Your motivating example is incorrect when $n=1.$ $\endgroup$ – DanielWainfleet Jun 10 '18 at 3:49
  • 1
    $\begingroup$ @DanielWainfleet No it's not, but even if it were it would be pretty useless to point out anyway. $\endgroup$ – 3-in-441 Jun 10 '18 at 3:53

Your Answer

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

Browse other questions tagged or ask your own question.