Proving any perfect disconnected subset is homeomorphic to Cantor Set

Question

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.

1. 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?

2. 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.

3. 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?

Answer 1

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.