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My comment reposted as an answer: If the continuum hypothesis holds, then we can give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ w …

Here is a more concrete construction. Let $X=\left\{\frac{n}{2^m};n\in\mathbb{Z},m\in\mathbb{N}\right\}\subseteq\mathbb{Q}$ and consider the set $$S=(\mathbb{R}\setminus\mathbb{Q})^2\cup\{(x,y)\in\mat …

It seems an path-connected anti-convex subset of $\mathbb{R}^2$ containing $(\mathbb{R}\setminus\mathbb{Q})^2$ exists. Firstly, let $A$ be a countable, dense subset of $\mathbb{R}^2$, and let $B$ be t …

Apparently not. Let $\gamma=1/3$ and choose some $\varepsilon<\frac{1}{4}$. For any $k\in\mathbb{N}$ we can consider the set $$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\;\sum_{i=1}^k\lfloor2x_i\rfloor\equiv …

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_ …

Yes, $\mathbb{R}^n\setminus E$ has to be path-connected. Let $x,y\in\mathbb{R}^n\setminus E$, we will prove that there are many paths going from $x$ to $y$ inside $\mathbb{R}^n\setminus E$. We can sup …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

We can first express the open set $U$ as a countable union of basic clopen subsets $A_n=\{(q_k)_{k\in\mathbb{N}};i_{n,k}<q_k<j_{n,k} \text{ for $k=1,\dots,n$}\}$, where $i_{n,k}$ and $j_{n,k}$ are irr …

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 …