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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)= …

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 $\mathbb Q$ …

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. There are two (discontinuous) surjective maps $f,g:S^1\to S^1$ whose graphs are connected but the graph of $g\circ f$ (as well as its closure) is not. The map $f$ is defined as follows, using th …

The comments explain how to prove the fact. If you want to put a formal wrapping around it, consider the strong (Whitney) $C^\infty$ topology on the space of maps $W\to M$. The strong $C^0$ topology …

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 …

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 …

Here is a formalization of André Henriques' answer to the Hausdorff dimension variant of the question. Let $K=\{0,1\}^\infty$ be the standard Cantor set. Define a map $f:K\to[0,1]$ as follows: for 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 …

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 …

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 …

No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces). Bad example: $X=\{a,b,c\}$ with open sets $\emptyset$, $X$, $\{a\}$, $ …

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( …