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That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$? I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t b …

It seems there aren't any counterexamples, even if $f$ is homogeneous but not even. If there is some counterexample $f$ to the question, and letting $X=\mathbb{R}^3\setminus\{x_1=x_2=x_3\}$, we can co …

It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also …

There are $2^{\aleph_0}$ different subsets of the Cantor set up to homeomorphism. There can't be more than $2^{\aleph_0}$ of them because any subset of the Cantor set is separable. To construct $2^{\a …

Too long for a comment. In the case $n=2$ and if the $U_i$ are connected, we can prove that $U_1\cap U_2\cap U_3$ is nonempty if the $U_i\cap U_j$ are nonempty: letting $p_1,p_2,p_3$ be in $U_2\cap U_ …

This doesn't work if the $U_\lambda$ are not connected, for example we can take $U_1=(-\infty,0)\cup(1,\infty)$, $U_2=(-1,1)$ and $U_3=(0,2)$ and the functions $f_1$ defined by $f_1(x)=x$ if $x<0$ and …