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Q1: No. Let $g(0)=1, g(1)=0$ and $g(x)=x$ for all $x\in\mathbb R\setminus\{0,1\}$. Assuming $f\circ f=g$, let $a=f(0)$, then $f(a)=1$ and $f(1)=g(a)=a$ since $a\notin\{0,1\}$. Then $g(1)=f(f(1))=f(a)= …

Enumerate all rational points outside your set. Then cover these points by open balls by induction as follows: the next ball is centered at the first rational point not covered so far, its radius is s …

For $A=[0,1]$, let $B$ be the 2-torus with one hole and $C$ be the 2-disc with two holes. The products $B\times[0,1]$ and $C\times[0,1]$ can be realized in $\mathbb R^3$: the former as a thickening o …

Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d-1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The la …

Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$ …

A complete space without isolated points has at least continuum cardinality. At least if you agree to use (some form of) Axiom of Choice. Choose two disjoint closed balls $B_1$ and $B_2$. Inside $B_1 …

This is not a reference but a short direct proof. Let $\bar X$ be the completion of $X$. Define $f:X\to\mathbb R$ by $f(x)=dist(x,\bar X\setminus X)$. Obviously $f$ is continuous, and the local compl …

This is not true even in finite dimensions. There exists a decreasing sequence of complete Riemannian metrics on the plane, pairwise Lipschitz equivalent, such that the pointwise limit is isometric to …

This is possible even on the real line. There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor …

No, it is not even path-connected in general, already for $n=1$. Consider the folding map $f:[0,1]\to[0,1]$, namely $f(t)=2t$ for $t\le 1/2$ and $f(t)=2(1-t)$ for $t\ge 1/2$. There is no path connect …

Here is another example which is easy to visualize. Let $X=[-1,1]^2$, $I$ be the segment between $(-1,0)$ and $(1,0)$ and $I_0$ its subsegment between $(-1/2,0)$ and $(1/2,0)$. There is a homeomorphis …

I think what you are looking for is boundary at infinity. For example, the boundary at infinity of the hyperbolic plane $\mathbb H^2$ is its "ideal boundary" circle, and adding it to $\mathbb H^2$ yie …

No. Let $B=S^1$, $F=\mathbb Z$, $E=S^1\times\mathbb Z$, and let $p$ be a two-fold covering on each component: $p(z,n)=z^2$ where $S^1$ is regarded as the unit circle in $\mathbb C$ (for the purposes o …

Yes. Let $A$ be the set in question. We may assume that $0\notin A$ and moreover that $A$ is outside the unit ball centered at the origin. Since $A$ is closed (in a complete space) and not compact, i …

The space of all functions $\mathbb R\to\mathbb R$ with the topology of pointwise convergence is obviously sequential but not first countable. Indeed, for every $x\in\mathbb R$, the set $U_x:=\{f:|f( …