A problem dealing with homeomorphism of metric spaces .

Question

$\mathbf {The \ Problem \ is}:$ Does there exist any $n>1$ the two metric spaces $C[0,1]$ under the uniform metric and $\mathbb R^n$ under the Euclidean metric, homeomorphic ???

Actually, I was thinking about this problem during proving path-connectedness of $C[0,1]$. Of couse, it is . And as the degrees of freedom on $C[0,1]$ is much more than that in $\mathbb R$, then they can't be homeomorphic .

But, removing countably many points from both $\mathbb R^n$ for $n>1$, and $C[0,1]$ makes them still path-connected .

But, thinking in an eerie sense, as a vector space, $C[0,1]$ is infinite dimensional, but $\mathbb R^n$'s are not for any $n$ and $\mathbb R^m$ and $\mathbb R^n$ are not homeomorphic for any $m>n$, so in that sense it seems that the statement might be false .

$\mathbf {My \ approach} :$ I thought the other way, what if I could disprove the statement ???

First of all, $card(C[0,1]) = card(\mathbb R^n) = 2^{\aleph_0}$ .

Secondly, both are seperable, the former is due to Stone-Weierstrass theorem and the latter due to $\mathbb Q^n$ .

Thirdly, neither is totally bounded, the former one has a subset of all constant functions of positive integers $\{1,2,3,4,....\}$ and the latter one due to non-compactness .

Now, I was thinking to prove the statement true by considering the fact of decimal representation of each point in $\mathbb R^n$, $n>1$ , but I don't find any assignment, even at least for continuity .

Now, I need serious help !!!

Answer 1

In $\mathbb R^n$, there are compact sets whose interior is not empty. This doesn't happen in $\mathcal C[0,1]$, since there no closed ball is compact. In fact, since $\mathcal C[0,1]$ is infinite-dimensional and its metric is induced by the $\sup$ norm, you know that the closed unit ball is not compact, which is equivalent to the assertion that no closed ball is compact. But then the interior of any compact subset of $\mathcal C[0,1]$ is empty, since, if $K\subset\mathcal C[0,1]$ is compact and if $f\in\mathring K$, then there is a $r>0$ such that $B(f,r)\subset K$, which implies that $\overline{B(f,r)}\subset\overline K=K$. But this is impossible, since then $\overline{B(f,r)}$ would be compact too, and we have already seen that that does not occur.