I'm a having some trouble understanding the proof of this theorem, the proof is taken from Charles Pugh Mathematical analysis.
The proof basically has two parts, one is the construction of the function and the second is proving that it is onto and continuous. Here $C$ denotes the standard cantor set(middle thirds).
Construction:
Step 1:
First we construct a function between the set of addresses and the cantor set. Take the code $0 =$ left and $2 = $ right.
So $C_0 = $ left interval = $[0, 1/3]$ and $C_2 = $ right interval $= [2/3, 1]$
$C_{00}$ would be the left subinterval of the left interval and $C_{01}$ would be the right subinterval of the left interval and so on.
let $\omega$ be an infinite address string $\omega = \omega_1 \omega_2 \omega_3 ...$ of zeroes and twos. We form a nested sequence of sets $$ C_{\omega_1} \supseteq C_{\omega_1\omega_2} \supseteq \dots \supseteq C_{\omega_1\omega_2\dots \omega_n} \supseteq \dots $$
Since these sets are nested, nonempty, compact and the diameter of each set goes to $0$, its infinite intersection is a point $p = p(\omega)$. $$ p(\omega) = \bigcap_{n \in N} C_{\omega | n} $$ where $\omega | n = \omega_1...\omega_n$ truncates $\omega$ to an address of length $n$. So each infinite defines a point in the cantor set and each point in the cantor set has an address.
Step 2:
Lemma: If $M$ is a nonempty compact metric space and $\epsilon > 0$ is given then $M$ can be expressed as the finite union of pieces, each of diameter $\leq \epsilon$. where a piece of $M$ is any compact nonempty subset of $M$
We say that $M$ divides into these small pieces. The metaphor is imperfect because the pieces may overlap. The strategy of the proof is to divide $M$ into large pieces, divide the large pieces into small pieces, divide the small pieces into smaller pieces and continue indefinitely. Labeling the pieces coherently with words in two letters leads to the Cantor surjection.
Let $W(n)$ be the set of words of two letters, say $a$ and $b$ having length $n$. Then #$W(n) = 2^n$
Using the Lemma above we divide $M$ into a finite number of pieces of diameter $\leq 1$ and we denote by $\mathcal M_1$ the collection of these pieces. We choose $n_1$ with $2^{n_1} \geq$ #$\mathcal M_1$ and choose any surjection $w_1 : W(n_1) \to \mathcal M_1$. Since there are enough words in $W(n_1)$, $w_1$ exists. We say $w_1$ labels $\mathcal M_1$ and if $w_1(\alpha) = L$ then $\alpha$ is a label of $L$.
Then we divide each $L \in \mathcal M_1$ into finitely many smaller pieces. Let $\mathcal M_2(L)$ be the collection of these smaller pieces and let $$ \mathcal M_2 = \bigcup_{L \in \mathcal M_1} \mathcal M_2(L)$$
Choose $n_2$ such that $2^{n_2}\geq \max\{\text{#}\mathcal M_2(L) : L \in \mathcal M_1\}$ and label $\mathcal M_2 $with words $\alpha\beta \in W(n_1 + n_2)$ such that
$$ \text{If } L = w_1(\alpha) \text{ then } \alpha\beta \text{ labels the pieces } S \in \mathcal M_2(L) \text{ as } \beta \text{ varies in } W(n_2).$$
This labeling amounts to a surjection $w_2 : W(n_1 + n_2) \to \mathcal M_2$ that is coherent with $w_1$ in the sense that $\beta \mapsto w_2(\alpha\beta)$ labels the pieces $S \in w_1(\alpha)$. Since there are enough words in $W(n_2)$, $w_2$ exists.
Proceeding by induction we get finer and finer divisions of $M$ coherently labeled with longer and longer words. More precisely there is a sequence of divisions $(\mathcal M_k)$ and surjections $w_k : W_k = W(n_1 + · · · + n_k) \to \mathcal M_k$ such that
(a) The maximum diameter of the pieces $L \in \mathcal M_k$ tends to zero as $k \to \infty$.
(b) $\mathcal M_{k+1} $ refines $\mathcal M_k$ in the sense that each $S \in \mathcal M_{k+1}$ is contained in some $L \in \mathcal M_k$. (“The small pieces $S$ are contained in the large pieces $L$.”)
(c) If $L \in \mathcal M_k \text{ and } \mathcal M_{k+1}(L) $ denotes $\{S \in \mathcal M_{k+1} : S \subseteq L\}$ then $$ L = \bigcup _{S \in M_{k+1}(L)} S$$
The labelings $w_k$ are coherent in the sense that if $w_k(\alpha) = L \in \mathcal M_k$ then $\beta \mapsto w_{k+1}(\alpha\beta)$ labels $\mathcal M_{k+1}(L)$ as $\beta $ varies in $W(n_{k+1})$.
Step 3:
We are given a nonempty compact metric space $M$ and we seek a continuous surjection $σ : C \to M$ where $C $ is the standard Cantor set. $C = \bigcap C^n$ where $C^n$ is the disjoint union of $2^n$ closed intervals of length $1/3^n$.
For $k = 1, 2, . . .$ let $\mathcal M_k$ be the fine divisions of $M$ constructed above, coherently labeled by $w_k$. They obey (a)-(d). Given $p \in C$ we look at the nested sequence of pieces $L_k(p) \in \mathcal M_k$ such that $L_k(p) = w_k(ω|(n_1 + · · · + n_k))$ where $ω = ω(p)$. That is, we truncate $ω(p)$ to its first $n_1 +· · ·+n_k$ letters and look at the piece in $\mathcal M_k$ with this label. (We replace the letters $0$ and $2$ with $a$ and $b$.) Then $(L_k(p))$ is a nested decreasing sequence of nonempty compact sets whose diameters tend to $0$ as $k\to \infty $. Thus $\bigcap L_k(p)$ is a well defined point in $M$ and we set $$ \sigma (p) = \bigcap_{n \in N} L_k(p) $$
Proof of $\sigma$ is continuous:
If $p$,$ p' \in C$ are close together then for large $n$ the first $n$ entries of their addresses are equal. This implies that $σ(p)$ and $σ(p')$ belong to a common $L_k$ and $k$ is large.
Since the diameter of $L_k$ tends to $0$ as $k \to \infty$ we get continuity.
Question: I don't understand why this proves continuity, I dont see the relation between this and sequential continuity or epsilon-delta. let alone open sets.
Proof of $\sigma$ is onto:
Each $q \in M$ is the intersection of at least one nested sequence of pieces $L_k \in M_k$. For $q$ belongs to some piece $L_1 \in \mathcal M_1$, and it also belongs to some subpiece $L_2 \in \mathcal M_2(L_1)$, etc. Coherence of the labeling of the $\mathcal M_k$ implies that for each nested sequence $(L_k)$ there is an infinite word $α = α_1α_2α_3 . . . $such that $α_i ∈ W(n_i)$ and $L_k = w_k(α_1 . . . α_m)$ with $m = n_1 + · · · + n_k$. The point $p ∈ C$ with address $α$ is sent by $σ$ to $q$.
Question: Why do we need $m = n_1 +\dots + n_k$ ? shouldn't $m = k$? I don't get it.
And as a final question, in the text, Pugh writes that this theorem means that the cantor set is the universal compact metric spaces, of which all others are merely shadows. What does this mean? I don't get the intuition of this.
I personally prefer other ways to show the result, sketching:
Concrete step: there is a continuous onto map $f: C \to [0,1]$, where $C=\{0,1\}^\mathbb{N}$ is the Cantor set.
Step 1a: So there also is a map continuous $F$ from $C \simeq C^\mathbb{N}$ onto the cube $[0,1]^\mathbb{N}$.
Step 2: every compact metric space $M$ embeds into $[0,1]^\mathbb{N}$ (Tychonoff embedding theorem).
Fact: every closed subset of $C$ is a retract of $C$, because the Cantor cube is $\textrm{AE}(0)$.
Combining: the map onto $M$ (if we see $M$ as a subset of $[0,1]^\mathbb{N}$) is just the composition of the retract $C$ to $F^{-1}[M]$ and $F$, essentially. (with some homeomorphisms/embeddings inserted).