Prove that $S$ is not connected

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.

Show that if $S\subset X$ is a subspace with two points, then $S$ is not connected.

My attempt:

Let $S=\{(s_n),\ (t_n)\}$ where $(s_n),\ (t_n)\in X$.

$U=\{(s_n)\}$ and $V=\{(t_n)\}$ are the only candidates for a separation of $S$.

$U\cap V=\varnothing$ and $U\cup V=S$.

$U$ is open in $S$ iff $U=S\cap A$ where $A$ is open in $X$.

How do I find such an $A$?

Answer 1

$(s_n)$ and $(t_n)$ being distinct, they differ in at least one position. WLOG, assume $s_0=0$ and $t_0=1$.

$A=\{0\} \times X$ and $B=\{1\} \times X$ should do the trick.

Answer 2

If $S$ has two distinct ponts $(s_n)_{n\in\mathbb N}$ and $(t_n)_{n\in\mathbb N}$, then $s_n\neq t_n$, for some $n\in\mathbb N$. Suppose, for instance, that $s_3=0$, whereas $t_3=1$. Take the sets$$A=\{0,1\}\times\{0,1\}\times\{0\}\times\{0,1\}\times\{0,1\}\times\cdots$$and$$B=\{0,1\}\times\{0,1\}\times\{1\}\times\{0,1\}\times\{0,1\}\times\cdots.$$Both of them are open sets, they are disjoint, $(s_n)_{n\in\mathbb N}\in A\cap S$, and $(t_n)_{n\in\mathbb N}\in B\cap S$.

Answer 3

The product of Hausdorff spaces is Hausdorff, and a dicrete space is Hausdorff. So you can find two open sets $W,W'$ in $X$ such that $W\cap W'=\emptyset$, $(u_n)\in W$ and $(v_n)\in W'$. Now, $U=S\cap W$ and $V=S\cap W'$, so $U,V$ are open in $S$.

Answer 4

Let $n$ be an index such that $s_n\ne t_n$, then consider the open sets $$\pi_n^{-1}(0)\ \text{ and }\ \pi_n^{-1}(1)$$