Some regular open sets of the Cantor space $2^\omega$ are not closed. The Cantor space is the set $2^\omega$ equipped with the product topology (where $2=\{0,1\}$ has the discrete topology).
I'm encountering some difficulties in understanding this proof (splitted into a list by me):

Proposition. Some regular open sets of $2^\omega$ are not closed.
Proof. A counterexample is given by Odd and Even, where Odd (resp. Even) is the set of sequences in $2^\omega$ that differ from $0^\omega$ and start with an odd (resp. even) number of zeros. These two sets are open, disjoint and

*

*their closures intersect only in $0^\omega$.

*In particular, $0^\omega$ is the unique point in the closure of Odd and Even such that no open set containing it has a dense intersection with Odd or a dense intersection with Even.

*While any element of Odd (resp. Even) has a clopen neighboorhood fully contained in Odd (resp. Even).

*This means that Odd and Even are regular and open,

but they are not closed.

In particular, it seems that the proof makes use of some sort of "intuitive understanding" of how the Cantor space works, which I can't grasp. I know that open sets are basically the ones where only a finite number of "entries" are fixed, and that $2^\omega$ is compact, Hausdorff and $0$-dimensional. But I don't understand:

*

*How do closures of sets in $2^\omega$ look like in general? And in particular, what are the closures of Odd and Even?

*How do closed sets look like in $2^\omega$?

*How do clopen sets look like in $2^\omega$?

*Why is (1) interesting/needed for the rest of the proof?

*Why is (2) true?

I understand why (3) is true and also why (4) follows by (3) (it's a characterization of regular open sets, even though we actually need just open neighborhoods, not clopen).
Maybe everything becomes more clear if we look at $2^\omega$ as the metric space with the well-known metric? I would be happy if someone could answer my questions and/or give any reference where I can learn how to better figure the Cantor space.
 A: It’s not really possible to give a general description of closures that’s significantly simpler than the definition of closure. 
I’ll use $O$ and $E$ for your sets $Odd$ and $Even$. Suppose that $x=\langle x_n:n\in\omega\rangle\in\operatorname{cl}O$. For each $n\in\omega$ let $B_n(x)=\{y\in 2^\omega:y_k=x_k\text{ for all }k\le n\}$; then $\{B_n(x):n\in\omega\}$ is a local base (of clopen sets) at $x$. Thus, for each $n\in\omega$ we have $B_n(x)\cap O\ne\varnothing$. If $x\ne 0^\omega$, let $m\in\omega$ be minimal such that $x_m=1$; since $B_m(x)\cap O\ne\varnothing$, there is a $y\in O$ such that $y_k=x_k$ for all $k\le m$, which means that $m$ must be odd, and therefore $x\in O$. Could $x$ be $0^\omega$? Yes: for each $n\in\omega$ we can find a $y\in O$ that begins with more than $n+1$ zeroes, and then $y\in B_n(0^\omega)\cap O$. We’ve just proved that $\operatorname{cl}O=O\cup\{0^\omega\}$.
Essentially the same argument shows that $\operatorname{cl}E=E\cup\{0^\omega\}$. Thus, neither $O$ nor $E$ is closed, and hence neither is clopen. Moreover, $(\operatorname{cl}O)\cap\operatorname{cl}E=\{0^\omega\}$. To complete the proof, we must show that $O$ and $E$ are not just open, but regular open. This is where (2) is supposed to come in, but I think that it’s unnecessarily confusing, so I’ll do it differently.
To show that $O$ is regular open, for instance, we must show that $\operatorname{int}\operatorname{cl}O=O$, i.e., that 
$$\operatorname{int}\big(O\cup\{0^\omega\}\big)=O\;.$$
Since $O$ is open, this amounts to showing that $0^\omega\notin\operatorname{int}\big(O\cup\{0^\omega\}\big)$. But we just saw that for each $n\in\omega$ there are points $y\in B_n(0^\omega)\cap O$ and $z\in B_n(0^\omega)\cap E$, and $O\cap E=\varnothing$, so $0^\omega$ has no open nbhd contained in either $O$ or $E$; in other words, $0^\omega\notin\operatorname{int}\big(O\cup\{0^\omega\}\big)$ and $0^\omega\notin\operatorname{int}\big(E\cup\{0^\omega\}\big)$, so $\operatorname{int}\big(O\cup\{0^\omega\}\big)=O$, $\operatorname{int}\big(E\cup\{0^\omega\}\big)=E$, and $O$ and $E$ are both regular open.
The actual statement of (2) is unnecessarily opaque in my opinion; let me see if I can clear it up a bit. Suppose that $0^\omega$ had an open nbhd $U$ such that $O\cap U$ was dense in $U$. Then 
$$U\subseteq\operatorname{cl}U=\operatorname{cl}(O\cap U)\subseteq\operatorname{cl}O=O\cup\{0^\omega\}=2^\omega\setminus E\;,$$
contradicting the fact that $0^\omega\in\operatorname{cl}E$. A similar contradiction would arise if $0^\omega$ had an open nbhd $U$ such that $E\cap U$ was dense in $U$. All of this is really just an unnecessarily roundabout way of demonstrating that $0^\omega$ has no open nbhd contained in $O\cup\{0^\omega\}$ or $E\cup\{0^\omega\}$ and hence that it’s not in the interior of either of these sets.
Closed sets are simply the complements of open sets, and open sets are the sets that are the unions of basic open sets of the form $B_n(x)$ for $n\in\omega$ and $x\in 2^\omega$. It’s not much easier to pin down the clopen sets.
A: Here's my intuitive picture of closures in $2^\omega$:
First, an exercise. A tree is a subset of $2^{<\omega}$ - that is, a set of finite binary sequences - which is closed downwards (if $\sigma\in T$ and $\tau\prec\sigma$, then $\tau\in T$). A path through a tree $T$ is an $f\in 2^\omega$ such that every initial segment of $f$ is in $T$. For instance, if $T$ is the set of all finite binary strings which don't contain "1" - so, $T=\{0, 00, 000, . . . 
\}$ - then the string "$00000. . .$ " is a path through $T$.
Exercise. If $X$ is the set of paths through some tree $T$, then $X$ is closed.
Now let's talk about closures. Given any set $X\subseteq 2^\omega$, there is a tree $T_X$ associated to $X$: $$T_X=\{\sigma: \exists f\in X(\sigma\prec f)\},$$ that is, $T_X$ is the tree of all initial segments of elements of $X$. Then it turns out (exercise) that the set $Y$ of paths through $T_X$ is exactly the closure of $X$.
For example, let $X$ be the set of all infinite binary strings which contain only finitely many "$0$"s. Then $T_X$ is all of $2^{<\omega}$: given any finite binary string $\sigma$, the infinite binary sequence $$\sigma111111111 . . .$$ is in $X$; so by definition of $T_X$, each finite binary string $\sigma$ is in $T_X$. But then the set of paths through $T_X$ is all of $2^\omega$.
I find trees easy to picture and draw examples of - even though they're infinite objects, they're not too nasty. By the way, if you prove the bolded exercise above, you'll have shown:
Corollary: A set $X\subseteq 2^\omega$ is closed iff $X$ is the set of paths through some tree.
And conversely, a set $Y$ is open iff $Y$ is the set of infinite binary sequences which are not paths through some tree.
