Cantor set - is it made of $[a,b]$ intervals or exclusively of singletons? I can't figure out whether the Cantor set $P = \bigcap E_n$ is made of very small intervals of the form $[a,b]$ ($a\neq b$) or whether it is made of uncountable many singletons ?
The first hypothesis tends to be confirmed by the fact that $P$ contains not only the end points of "remaining intervals" but also extra points (such as $\frac{1}{4}$ or $\frac{3}{4}$).
The second hypothesis tends to be confirmed by this reasoning: "no interval $[a,b]$ of length strictly greater than $\frac{1}{3^m}$ is included in $E_m$. So choosing $m$ big enough so that $b-a >\frac{1}{3^m}$ we have $[a,b] \not\subset E_m$. Hence, $[a,b] \not\subset P$ for $a\neq b$"
 A: When you say

I can't figure out whether…

that's good!  It means you apprehend why the Cantor set is considered pathological and is a famous counterexample:  it's weird.  Sometimes I meet people who say

I don't see what is so strange about the Cantor set

or even

There is nothing strange about the Cantor set

and you're a step farther along than those people.
Your second argument is fine.  There are no intervals contained in the Cantor set:  Any interval $(a,b)$ includes a point $c$ for which $c\notin P$, so therefore $(a, b)\not
\subset P$.
Your idea about "uncountable many singletons" is hard for me to understand though.  A singleton is just a set with one element $\{x\}$.  And every set is a union of singletons:
$$S = \bigcup_{x\in S} \{x\}.$$
So I think you certainly don't mean "singleton", but I'm not sure what you do mean. Maybe something like "isolated point"? The Cantor set has no isolated points though.
It might help to think here about the rational numbers $\Bbb Q$ or the irrationals $\Bbb R - \Bbb Q$.  Like the Cantor set, these sets contain no intervals.  Would you say it they are unions of "singletons", whatever you mean by that?  (Both are of course unions of singletons since every set is a union of singletons, but what I want to know if they have the property you had in mind when you said singletons.)
The irrationals are another uncountable set that contains no intervals and no isolated points; both  $\Bbb R - \Bbb Q$ and $P$ are what's called totally disconnected.  The Cantor set, though, is closed, which the set of irrationals is not.  (The proof is trivial: the Cantor set is closed because it's an intersection of closed sets.)

Update: You asked in a comment:

Is there an $n$ such that $\frac 14\in E_n$ ? Or $\frac 14\in P$ without belonging to any $E_n$ ?

This is confused. By definition $$P = \bigcap E_n$$ and therefore for $\frac 14$ to be in $P$, it must be in every $E_n$.  If it were missing from any $E_n$, it would not be in their intersection.


*$E_0$ is $[0,1]$, and certainly $\frac14 \in[0,1]$.


*$E_1$ is $[0,\frac13]\cup [\frac23, 1]$, and $\frac 14$ is in the left part, $[0,\frac13]$.


*$E_2$ is $[0,\frac19]\cup [\frac29, \frac13]\cup\dots$, and $\frac 14$ is in the $[\frac29, \frac13]$ part, because $\frac29\le\frac14\le\frac 13$.
You might want to think about the next one, and then do to a similar analysis for $\frac15$.  The question to ask about $\frac 15$ is “which $E_n$ is it not in?”
