As Andre Nicolas commented, if we only remove the end points we are left with more or less the entire Cantor set in terms of cardinality. Topologically speaking, however, this is not the case.
A very interesting thing happens when removing these end points, though. I will elaborate on that in a short while. First we introduce a few concepts (and while at it take this question from real-analysis into descriptive-set-theory and general-topology)
First I want to introduce the Baire space, denoted often by $\omega^\omega$ and $\mathcal N$. This is the space of infinite sequences of natural numbers, but we can look at this space as though it were a tree.
In the root we have the empty sequence, on the first level we have sequences in length one, that is $\langle n\rangle$, then we have sequences in length of two, three, and so on. Each vertex in the tree has infinitely many immediate successors. The Baire space is the space formed when the tree has infinitely many levels, and we consider the branches (and not the vertices) as the elements.
To compare, the Cantor set has a similar construction when considering the underlying set $\{0,1\}$ as Dylan showed in his answer with $3$-adic expansions limited to digits $0,2$ in the expansion.
The Baire space (and every tree of this form as well) can be endowed with a topology - that is we say which sets are open and which are closed. In this case we let a basic open set (a ball of some positive radius, if you will) to be all branches that share the same initial segment of a prescribed length.
Examples would be all the sequences in which the first ten coordinates are $42$, or the sequences that in the first six coordinates are $\langle 2,574,87144781, 2^{1000}, 314159\rangle$.
We can also define a metric function on the Baire space, turning it into a metric space (similar to $\mathbb R$, yet very very different). If we have two distinct branches, they must have at least one $n$ in which they are different. Therefore they have a minimal $n$ with this property, so we say that they are in distance of $2^{-n}$. Formally:
$$d(u,v) = \begin{cases}
2^{-n} & \text{ where } n = \min\{k\mid u(k)\neq v(k)\} \\
0 &\text{ otherwise}
\end{cases}$$
This is an ultrametric, which means that every element inside an open ball is its center (kind of weird, compare with $p$-adic numbers).
A very interesting fact about the Cantor set is that it is the only (up to homeomorphism) space which is totally disconnected, separable, has no isolated points and compact.
Two interesting facts about the Baire space:
It is homeomorphic to the irrationals, that is $\mathbb R\setminus\mathbb Q$, with the subspace topology. It is also a complete metric space. While this seems somewhat strange, since clearly the irrationals are incomplete (take a sequence going to $0$). However the two metrics are only topologically equivalent, this means that if we were to measure distance in a way similar to the way we do in the Baire space, a sequence of irrationals will converge to an irrational limit.
This is a totally disconnected space, without isolated points, it has a countable dense subset, and every compact subset of the Baire space has an empty interior. To compare the Cantor set has an empty interior in the reals, that is there is no open interval contained within the Cantor set. To some extent this means that compact subspaces of the Baire space "look" a bit like the Cantor set (to be fair, they look exactly like the Cantor space, but this is not the point here).
Like the Cantor set, the Baire space is unique (up to homeomorphism, of course) space with this (the second) property. Which makes both the Cantor set and the Baire space very interesting in the field of descriptive set theory.
Now to the main punch, suppose we take the Cantor set, and remove all the end points of the intervals (i.e. we remove closed intervals in the construction) the result is a space which zero dimensional, totally disconnected, without isolated points, in which every compact subspace has an empty interior!
That means that we start with the Cantor set, removed a relatively small collection of points and ended up with the irrational numbers, which is a completely different space.
As for the compact notation for $C_n$, note that every stage you remove more and more intervals. In fact at the $n$-th stage we remove $2^n$ intervals (as well all that you removed before). The Wikipedia page suggests the following notation:
$$C_n = [0,1]\setminus\bigcup_{k=0}^{2^m-1} (\frac{3k+1}{3^m},\frac{3k+2}{3^m})$$
Note that in this notation the $C_n$ are not a decreasing sequence of sets, if you want this property as well replace $[0,1]$ by $C_{n-1}$ (and let $C_0 = [0,1]$ of course).
I begin by assuming that by the $C_n$ you mean the usual closed sets whose intersection is the Cantor ternary set:
- $C_1 = [0,1]$;
- $C_2 = [0,\frac 13] \cup [ \frac 23 , 1 ]$;
- $C_3 = [0,\frac 19] \cup [\frac 29,\frac 13] \cup [\frac 23,\frac 79] \cup [\frac 89, 1]$;
- etc.
Note that for each $n$ the set $C_n$ is made up of disjoint closed intervals of length $3^{-(n-1)}$, and that these intervals are therefore separated from each other. Also note that if $x,y \in C_n$ are such that $3^{-(n-1)} < |x-y|$, then $x,y$ belong to different closed intervals making up $C_n$.
Given distinct $x,y \in C$ essentially by the Archimedean property there must be an $n$ such that $3^{-(n-1)} < |x-y|$, and as $x,y \in C_n$ it follows that they belong to different closed intervals making up $C_n$. Let $I$ be the closed interval in $C_n$ containing $x$. It follows that $x \in C \cap I$ and $y \in C \setminus I$, and these sets are separated.
Best Answer
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.