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This answer was supposed to contain a counterexample, but there was a mistake in it. In any case I think the following class of topological spaces may be useful to find a counterexample: Suppose that …

Extending $\phi$ to the ball is equivalent to proving that $\phi:S^{k-1}\to H\setminus Q$ is nulhomotopic. To prove this, you can consider the map $F:\phi(S^{k-1})\times Q\to E;(x,y)\to\frac{y-x}{|y-x …

The answer to the second question is yes: there is an arc containing uncountable points of $B\cap\partial C$. It is enough to prove it in the case $n=2$. Applying an affine transformation if necessary …

Calling the space $X$, you can consider the homotopy $f_t:X\to X$ such that $f_t$ rotates every point an angle of $-t$ around the origin. For points of the hairs you rotate them keeping them inside th …

There are lots of classes of RSTs (rational sequence topologies) in $\mathbb{R}$ up to homeomorphism. Note that, as mentioned in the answer by Will Brian, any homeomorphism $(\mathbb{R},T_1)\to(\mathb …

Here is one way to form a (not very) explicit homeomorphism. As has been mentioned, the function $\phi=(\phi_0,\phi_1)$ from the question applied to the set $I^{>1}$ of irrationals greater than $1$ gi …

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 …

For a loose metric $d$ as above, we can consider the function $$d_1(x,y):=\sup\{|d(x,z)-d(y,z)|;z\in X\}.$$ It is easy to verify that $d_1$ is a metric, and $d(x,y)\leq d_1(x,y)\leq\rho(d(x,y))$ for a …

Yes, such a distance $d'$ exists. We can suppose $d(x,X)=d(y,X)=k$ for some $k>0$. We can define a new distance $d'$ by $d'(a,b)=d(a,b)+|d(x,a)-d(x,b)|$. This easily implies $d'(x,X)=2k$, however for …

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 question is answered positively in On closed images of the space of irrationals, by Engelking.

It seems like there are decomposable rays which limit onto themselves. Let $E=\bigcup_{n\geq0}E_n$ and $O=\bigcup_{n\geq0}O_n$, with $E_n=[2n,2n+1]$ and $O_n=[2n+1,2n+2]$. We want to construct a ray $ …

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 …

A convex $O'$ need not exist: a counterexample is given by setting $K=[-1,1]\times[0,2]\subseteq\mathbb{R}^2$ and $O=\{(x,y)\in K;y>x^3\}$. Indeed, any open $O'$ with $O'\cap K=O$ would contain some n …

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 …