Product topology (infinite product)

Question

Consider the two-element set $\{0,\ 1\}$ equipped with the discrete topology, and form the countably infinite product

$\displaystyle X:=\{0,\ 1\}^\omega=\prod\limits_{n\in\mathbb{Z_+}}\{0,\ 1\}$

So $X$ consists of the infinite sequences $\displaystyle(x_n)_{n\in\mathbb{Z_+}}$, where for each $k\in\mathbb{Z}_+$, the $k$th term $x_k$ is either $0$ or $1$. Equip $X$ with the product topology.

Let $(s_n)\in X$. Prove that $\{(s_n)\}$ is open in $X$.

My attempt:

My hunch is that since $\{0,\ 1\}$ is equipped with the discrete topology, and $X$ is equipped with the product topology, this amounts to equipping $X$ with the discrete topology as well. In that case, every subset of $X$ would be open. However, I haven't learn this result in class and so can't use it unless I prove it.

Answer 1

This is not true. It would be that the topology of $X$ is the discrete topology, but it isn't. For instance, by Tychonoff's theorem $X$ is compact, but an infinite discrete topological space is never compact.

Answer 2

The statement you are trying to prove is wrong. The space you are describing is homeomorphic with the Cantor set (see https://en.wikipedia.org/wiki/Cantor_space) which has no isolated points and hence no open singleton sets.

Alternatively, you can think of it like this: the product topology on $X = \{0, 1\}^{\omega}$ has a basis comprising sets of the form $\{\langle s_1, \ldots, s_k\rangle\} \times X$. You can think of this topology as a topology of "finite information": given an open set $U$, to decide whether $(s_n) \in U$ you only need to test finitely many of the $s_n$. So $\{(s_n)\}$ can't be open since the assertion $(x_n) \in \{(s_n)\}$ gives information about $x_n$ for all $n$.