How does the Cantor set contain more points than the boundary points of those deleted open intervals? I'm trying to get an intuitive grasp of the Cantor set.
Let $$C_0 = [0,1]$$ and $$C_n = \frac{C_{n-1}}{3} \cup (\frac{2}{3} + \frac{C_{n-1}}{3}).$$ Then we call $$\mathcal C = \bigcap_{n \in \mathbb N}$$ the Cantor set.
Here's something that my intuition cannot resolve: The Cantor set is constructed in countably many steps where in each step a finite number of open intervals is deleted.
Now, it's quite clear that all the end points of those deleted intervals will remain in the set. However, these are just countably many since we perform a countable number of steps and each step only adds a finite number of end points.
Now, here's where my intuition escapes me: How is it that any point that is not an end point of one of these deleted open intervals remains in the set as well? Wouldn't any non-end-point be cut out eventually? How can I imagine this process properly?
 A: for example, $\frac{1}{4}$ is in the cantor set, even though it isn't in the boundary of any $C_n$.
To see why this is so, consider when $\frac{1}{4}\in C_n$. Since $\frac{1}{4}$ is smaller than a half, it can only be in the first half of $C_n$, so $\frac{1}{4}\in C_n$ iff $\frac{1}{4}\in \frac{C_{n-1}}{3}$, which is equivalent to $\frac 3 4 \in C_{n-1}$.
In the same way, $\frac 3 4$ can only be in the second half, so $\frac 3 4\in C_{n-1}$ iff $\frac 3 4\in \frac 2 3 + \frac{C_{n-2}}{3}$ which is equivalent to $\frac 1 4 \in C_{n-2}$.
Thus by induction $\frac 1 4$ is always inside $C_n$, and thus, it is also in their intersection, the cantor set.
A: This is hard to picture, because our mental images are based on the sets $C_n$, and $\mathcal C$ is the limit of these; our brains don't always deal well with infinity. Here's something that may help.
One way to think about the Cantor set is in terms of decimal expansions in base 3. Take any $x \in [0,1]$; written in base 3, we have $x = 0.a_1a_2a_3\ldots$, where $0 \leq a_i \leq 2$ for all $i$. Now convince yourself that the Cantor set consists precisely of those $x$ such that $a_i \in \{0,2\}$ for all $i$.
Under this identification, the points you are imagining as boundaries are exactly the points with a finite decimal expansion of this form, or which end with an infinite number of $2$s: e.g. $0.022222...$ corresponds to the point $1/3$ and $0.2$ corresponds to $2/3$. In the next step of the construction we get additionally $0.002222... = 1/9$, $0.02 = 2/9$, $0.202222... = 7/9$, and $0.22 = 8/9$. Clearly the number of points in the Cantor set of this form is countable.
Now try to imagine what points correspond to which are not of this form, e.g. $0.202002000200002\ldots$. Can you convince yourself that it's in the Cantor set? Can you convince yourself that it's not a boundary?
A: "Each step only adds a finite number of end points". No, each of those endpoints was already there in $C_0$, and each is there in $C_1$, and each is there in $C_2$, and so on.
You are thinking of what each step "adds" when instead you should be thinking of what each step "removes". What each step does is to remove all the points in a certain finite union of open intervals. Step 1 removes the points in $(1/3,2/3)$; step 2 removes the points in $(1/9,2/9) \cup (7/9,8/9)$; and so on. It's the points that are never removed that you should be focussing your attention on. 
The point $0$ is never removed, so it is in the Cantor set. Each point which ends up some time down the line being an endpoint of one of the removed intervals is itself never removed, and so each such point in the Cantor set. And there are many other points --- such as $\frac{1}{4}$ mentioned by @user3329719, or $0.202002000200002...$ mentioned by @user --- which are not endpoints and which nonetheless are never removed so each is in the Cantor set.
