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It's easy to see that, for $1\le p,q< \infty$ the spaces $L^p(\Bbb R)$ and $L^q(\Bbb R)$ of $p$-th and $q$-th power integrable functions on the real line are homeomorphic as topological spaces. In fact, the map $f(x)\mapsto sgn(f(x))|f(x)|^{q/p}$ provides an explicit homeomorphism $L^p(\Bbb R) \to L^q(\Bbb R)$.

However this argument cannot be applied to the case where $p$ or $q$ equals infinity. Thus, I'm asking whether there is a homeomorphism from $L^\infty(\Bbb R)$ to any (and hence every) $L^p(\Bbb R)$ with $p<\infty$.

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    $\begingroup$ Isn't $L^2(\mathbb{R})$ separable and $L^\infty(\mathbb{R})$ nonseparable? $\endgroup$
    – Paul McKenney
    Commented Feb 9, 2015 at 22:38
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    $\begingroup$ A more interesting question would be: are $L^2({\bf R})$ and $C_0({\bf R})$ homeomorphic? $\endgroup$
    – Yemon Choi
    Commented Feb 9, 2015 at 23:07
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    $\begingroup$ Any two separable infinite dimensional Frechet spaces are homeomorphic. This is due to M. I. Kadec. $\endgroup$
    – Bill Johnson
    Commented Feb 9, 2015 at 23:25
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    $\begingroup$ The $q/p$ should be $p/q$ if you want it to take $L^p$ to $L^q$. $\endgroup$
    – Robert Israel
    Commented Feb 10, 2015 at 2:24
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    $\begingroup$ @PaulMcKenney I like the use of the rhetorical question... $\endgroup$
    – Yemon Choi
    Commented Feb 10, 2015 at 12:18

1 Answer 1

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Paul is right. $L^2(\mathbb{R})$ is separable. (The rational simple functions ought to be one example of something ctbl. and dense.)

However, $L^{\infty}(\mathbb{R})$ isn't separable. By Jones' Lemma, if $L^{\infty}(\mathbb{R})$ were separable then any closed discrete (i.e., nonclustering) set must be of size less than continuum. But the characteristic functions $\chi_{[0,x]}$, $x\in[0,\infty)$, are pairwise of distance 1 from eachother.

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    $\begingroup$ This was already pointed out in comments $\endgroup$
    – Yemon Choi
    Commented Feb 10, 2015 at 9:54
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    $\begingroup$ And I was clearly elaborating on that comment. $\endgroup$
    – Michael Cotton
    Commented Feb 10, 2015 at 12:15
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    $\begingroup$ We need answers, since comments are likely to go away in the future. $\endgroup$
    – Gerald Edgar
    Commented Feb 10, 2015 at 13:07
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    $\begingroup$ We need answers, since questions without answers constantly float back to the first page. $\endgroup$
    – Steven Gubkin
    Commented Feb 10, 2015 at 14:29
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    $\begingroup$ Jones' Lemma seems like overkill here. Since the continuum-many functions $\chi_{[0,x]}$ are pairwise distance 1 in $L^\infty$, the continuum-many open balls $B(\chi_{[0,x]}, 1/2)$ are pairwise disjoint. Any dense set must contain points in every one of these balls, so a dense set in $L^\infty$ must have size at least continuum. $\endgroup$
    – Nate Eldredge
    Commented Jan 4, 2017 at 20:14

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