Proving any perfect disconnected subset is homeomorphic to Cantor Set I am reading this paper for a presentation and am trying to understand lemma 2.7 on page 9 to 10. Specifically, I understand the proof up till the last paragraph on page 9 and the top paragraph  on page 10.


*

*What does it mean the paper says "Since E and the Cantor Set are totally disconnected, they are nowhere dense and thus at most one continuation $F:[m, M] \rightarrow [0.1]$ ? I think the last part is saying that there exists at most one function that maps $F:[m, M] \rightarrow [0.1]$ that ensures a homeomorphism between the two spaces but what does it mean to be "nowhere dense" and what does it have to do with it?

*How does the author then arrive at the fact that $F(x) = sup(F(y) : y \notin E, y\leq x)$? I would write out what I do know about this but I have no idea as to how this conclusion is arrived at.

*On page 10, what is the contradiction that shows that the inverse function $g(x)$ is continuous. If I accept the $F(x)$ function and its properties from question 2 at face value then I understand that $f: E \rightarrow C$ is continuous and monotone increasing. But how exactly is $g: f^{-1}$ also continuous?
 A: There are several questions here; I'll answer (1) here. Others may address (2) and (3), but I think seeing the answer to (1) will give you some intuition for them which may help you answer them yourself.
A set $X\subseteq\mathbb{R}$ is nowhere dense if it's not dense in any open interval; formally, if for all nonempty intervals $(a, b)$ there is a nonempty subinterval $(c,d)\subseteq (a,b)$ such that $(c,d)\cap X=\emptyset$. Note that if $X$ is nowhere dense, then the complement of $X$ is dense. The converse is false, however - take $X=\mathbb{Q}$.
Now I claim that a totally disconnected compact set is nowhere dense. Actually, I claim that a totally disconnected closed set is nowhere dense (remember, compact sets in $\mathbb{R}$ are closed!). Why? Well, suppose $X$ is closed and not nowhere dense. Then for some nonempty interval $(a,b)$, $X$ is dense in $(a, b)$: for any $y\in (a,b)$ there are elements of $X$ arbitrarily close to $y$. (Exercise.) But since $X$ is closed, this means $(a,b)\subseteq X$, so $X$ is not totally disconnected. Conversely, if $X$ is closed and totally disconnected, $X$ must be nowhere dense.
As to the "at most one continuation" bit, all that's going on there is: if $g$ is a continuous map from $A\subseteq\mathbb{R}$ to $\mathbb{R}$, and $A$ is dense in $[0,1]$ (note: this follows if the complement of $A$ is nowhere dense), then there is at most one continuous $h:[0,1]\rightarrow\mathbb{R}$ extending $g$ (the point being that the value of $h$ on any element of $[0,1]\setminus A$ is already forced by the fact that $g$ is defined at points arbitrarily close to that element). The phrasing here is a bit weird.
