The topology of the Cantor Set For reference a cantor set and $C_n$ is described here https://en.wikipedia.org/wiki/Cantor_set : 
$C$ is a topological subspace of $[0,1]$, where $C= \cap C_n$, and $C_n$ is defined by removing the middle third open intervals from $C_{n-1}$ where $C_0 = [0,1]$.
Problem : 
Show that the sets obtained by intersecting the Cantor set $C$ with the interval components of the spaces $C_n$ form a basis for the topology on $C$.
First off, it seems to me that for $n < m$, $C_n \cap C_m$ = $C_m$. The wording of the question is leading me to be confused about what the basis elements actually are. I'm imagining they are of the from $C \cap C_n$, but then it would seem that $C \cap C_n=C$ itself, so this would be the indiscrete topology? Any insights appreciated.
 A: What is meant is that we use the ever-decreasing small subintervals from all the construction stages of the Cantor set:
Let's look at the construction, while introducing some handy notation (using standard ideas from descriptive set theory):
I'll denote by $\Sigma$ the set of all finite length sequences of $0$'s and $1$'s, like $\varepsilon$ (the empty sequence), $0$,$1$ (each of length $1$, I'll write $l(\sigma)$ for the length of a sequence,) and also $00,01,10,11$ etc. If $\sigma$ is a sequence from $\Sigma$, then $\sigma^\frown 0$ is that sequence $\sigma$ extended with a $0$ and likewise for $1$. If $\sigma$ is a sequence with $l(\sigma) \ge n$, I write $\sigma|_n$ for the sequence that is the first $n$ terms of $\sigma$, its restriction or initial sequence.
We start with $C_\varepsilon = [0,1]$ and we define $C_\sigma$ for a sequence by recursion on the length: if $C_\sigma$ has already been defined as some closed interval $[a,b]$, then we remove the middle third open interval from $[a,b]$ (i.e. $(a+\frac{b-a}{3}, b-\frac{b-a}{3})$ to get a left interval $C_{\sigma^\frown 0} = [a,a+\frac{b-a}{3}]$ and $C_{\sigma^\frown 1} = [b-\frac{b-a}{3},b]$. This allows us to define $C_\sigma$ with $l(\sigma) = n+1$ from the $C_\sigma$ with $l(\sigma)=n$, and we know what $C_\varepsilon$ is as our starting point. 
Define $K_n = \bigcup\{C_\sigma: l(\sigma)=n\}$, which is compact, as a finite ($2^n$ many) union of compact intervals. The Cantor set $C$ is defined as $C = \bigcap_n K_n$. E.g. $K_1 = C_0 \cup C_1 = [0,\frac13] \cup [\frac23, 1]$, $K_2 = C_{00} (= [0,\frac19]) \cup C_{01} (=[\frac29, \frac13]) \cup C_{10} (= [\frac23, \frac79]) \cup C_{11} (=[\frac89,1])$ and $C_{001}$ would be $[\frac{2}{27},\frac19]$, the right third of $C_{00}$ etc.
So if  $x \in C$, for each $n$,$x \in K_n$ so there is a unique $\sigma=\sigma_n(x)$ of length $n$ such that $x \in C_\sigma$. All those sequences are extensions of each other ($\sigma_n(x)|_{n-1} = \sigma_{n-1}(x)$ for all $ n \ge 1$) This defines a unique infinite sequence $h(x) \in \{0,1\}^\mathbb{N}$ by $h(x)|_n = \sigma_n(x)$ for all $n$.
And conversely, for each infinite sequence $\sigma \in \{0,1\}^\mathbb{N}$ we can define the compact sets $C_{\sigma|_n}$ which are nested and of decreasing diameter ($\operatorname{diam}(C_\sigma) = \frac{1}{3^{l(\sigma)}}$ for $\sigma \in \Sigma$ as follows by an induction argument) so Cantor's nested interval theorem implies that this intersection defines a unique point in $C$ (this is exactly the inverse map of the above $h$). So we get a bijection between $C$ and $\{0,1\}^\mathbb{N}$ which is an uncountable set by Cantor's theorem.  
Now, if $U$ is relatively open in $C$ and $x \in U \cap C$, by definition $U$ is of the form $O \cap C$ where $O$ is standard Euclidean open in the reals. So some $r>0$ exists so that $(x-r,x+r) \subseteq O$. Now let $n$ be large enough so that $\frac{1}{3^n} < r$, and let $\sigma$ be the unique sequence of length $n$ such that $x \in C_\sigma$. As $x$ lies in the interval, we have that $C_\sigma \subseteq (x-r,x+r)$ as well (any point of $C_\sigma$ lies within distance $\operatorname{diam}(C_\sigma)$ of $x$ so has distance $<r$ to $x$). It follows that $$x \in C \cap C_\sigma \subseteq (x-r,x+r)\cap C \subseteq O \cap C=U$$
and this shows that the set $$\{C \cap C_\sigma: \sigma \in \Sigma\}$$ obeys the condition of being a base for the subspace topology on $C$.
The only thing needed to check is that each $C \cap C_\sigma$ is not only closed in $C$ but also relatively open, and this follows as for each $\sigma \in \Sigma$
$$C\setminus (C \cap C_\sigma) = \bigcup \{C\cap C_{\sigma'}: \sigma' \in \Sigma, l(\sigma') = l(\sigma), \sigma'\neq \sigma\}$$ 
because the sets $C_\sigma$ for sequences of the same length $n$ form a closed partition of $K_n$ all $n$ (so each complement is relatively closed.)
With a little more work we can even show that the map $h$ defined above is a homeomorphism between $C$ and $\{0,1\}^\mathbb{N}$ and the $\sigma$ then define a simple base for the product topology on $\{0,1\}^\mathbb{N}$ namely $B_\sigma = \{x \in \{0,1\}^\mathbb{N}: x|_{l(\sigma)} = \sigma\}$ such that $h[C \cap C_\sigma] = B_\sigma$ etc. 
