7

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!

Bernard
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3-in-441
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  • Not that it answers the question, but X is already homeomorphic to its unit sphere without an added point; see here. – Michael Greinecker Jun 09 '18 at 23:02
  • @MichaelGreinecker Thank you for the link. That is interesting (I should have guessed that would be true since the sphere is contractible.) – 3-in-441 Jun 09 '18 at 23:46
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    Why wouldn't the stereographic projection work? – Berci Jun 10 '18 at 00:15
  • Your motivating example is incorrect when $n=1.$ – DanielWainfleet Jun 10 '18 at 03:49
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    @DanielWainfleet No it's not, but even if it were it would be pretty useless to point out anyway. – 3-in-441 Jun 10 '18 at 03:53
  • @Berci I thought I had an argument for why it didn't work, but I think I just made a mistake. I will write try to write it again tomorrow and post the answer either way. Thanks for the comment! – 3-in-441 Jun 10 '18 at 03:54
  • Yes, the example is incorrect - in fact it's a simple counterexample. Yes, if $X=\Bbb R^n$ then $Y\approx S^n$. But $S^n$ is not the unit sphere of $\Bbb R^n$, it is the unit sphere of $\Bbb R^{n+1}$. – David C. Ullrich Jun 10 '18 at 16:05
  • @DavidC.Ullrich I think you're both very clearly misreading the example. $S^n$ is obviously the $n$ dimensional unit sphere. I made no claim that $S^n$ sits in $R^n$. The question comes from the fact that it could be "$\infty + 1 = \infty$". – 3-in-441 Jun 10 '18 at 16:09
  • The question is whether $Y$ is homoemorphic to the unit sphere of $X$. The example $X=\Bbb R^n$ shows that this is not true. Because in that case $Y$ is homeomorphic to $S^n$, and $S^n$ is not homeomorphic to the unit sphere of $X$. – David C. Ullrich Jun 10 '18 at 16:22
  • @DavidC.Ullrich I stand by that the intent of the question is clear and quibbling about the exact wording is a complete of time and you're really detracting from the discussion. Do you have a proof, suggestion, or would you like to write out the stereographic projection as an answer or will you wait for me to? The entire point of math an this site is to communicate math, right? Stop wasting everyone's time. You're creating a toxic environment and you should be ashamed. – 3-in-441 Jun 10 '18 at 16:32
  • What are you talking about? Yes, the question is perfectly clear - I haven't been quibbling about the wording of anything. The question is clear, and $X=\Bbb R^n$ is a counterexample. – David C. Ullrich Jun 10 '18 at 16:36
  • @DavidC.Ullrich The first paragraph makes it clear that I am asking about when $X$ is infinite dimensional. – 3-in-441 Jun 10 '18 at 16:40
  • Yes come to think of it it does. I missed the last sentence. You might have simply pointed this out instead of being insulting. – David C. Ullrich Jun 10 '18 at 16:51
  • @DavidC.Ullrich I did in my first comment to you, although the way I phrased it probably made no sense if you were missing the context – 3-in-441 Jun 10 '18 at 16:54

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