6
$\begingroup$

I am somehow puzzled by the subset $G:=L^2(I,\mathbb Z)$ of $H:=L^2(I,\mathbb R)$ of all integer valued functions on $I=[0,1]$ (in fact I mentioned as an example in this old MO question).

Some simple observations: it is a complete metric topological additive group, though of course not a linear subspace; it is connected (even by $1/2$-Hölder arcs); it is contractible, via the homotopy $G\times[0,1]\ni(f,s)\mapsto h(f,s):=f\chi_{[0,s]}\in G$. And then, what else can be said?

Is it homeomorphic to the whole space $H$? Is there an explicit homeomorphism?

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ What would happen, if we compare $L^2(I,\{0,1\})$ with $L^2(I,\mathbb{R})$? I dont know if these spaces are homeomorphic. $\endgroup$
    – HenrikRüping
    Oct 19 at 14:30
  • 1
    $\begingroup$ The paths in the upper homotopy decrease the distances to $0$. Thus $\varepsilon$-Balls are contractible and hence $L^2(I,\mathbb{Z})$ is also locally contractible. $\endgroup$
    – HenrikRüping
    Oct 20 at 8:13

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.