Countable product of discrete sets

Hi I am having problems with this:

Let $X$ be a countable discrete topological space $\{0,1 \}$, I need to find a dense countable set in $X$, but don´t know which se could be work, thanks for the help.

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Do you really mean that $X$ is the two-point space $\{0,1\}$? – Brian M. Scott Oct 26 '12 at 15:05
Your title doesn't match your question. – Chris Eagle Oct 26 '12 at 15:07
Why the word product in the title? – Rudy the Reindeer Oct 26 '12 at 15:10
X is a countable product of discrete spaces Xi, where Xi={0,1} – Ali Oct 26 '12 at 15:12
@MattN.: Ali may have intended to ask about $\{0,1\}^\omega$ and expressed it very badly. – Brian M. Scott Oct 26 '12 at 15:12

HINT: Let $D=\{0,1\}$ with the discrete topology, for each $k\in\Bbb N$ let $D_k$ be a copy of $D$, and let $X=\prod_{k\in\Bbb N}D_k$.

For each $x\in X$ let $s(x)=\{k\in\Bbb N:x_k=1\}$, and let $A=\{x\in X:s(x)\text{ is finite}\}$; show that $A$ is a countable dense subset of $X$.

Added: Let $\Sigma$ be the set of finite sequences of $0$’s and $1$’s. For each $\sigma=\langle b_0,\dots,b_n\rangle\in\Sigma$ let $$B(\sigma)=\{x\in X:x_k=b_k\text{ for }k=0,\dots,n\}\;,$$ and let $\mathscr{B}=\{B(\sigma):\sigma\in\Sigma\}$; $\mathscr{B}$ is a base for the product topology on $X$, so to show that $A$ is dense in $X$, we need only show that $B\cap A\ne\varnothing$ for each $B\in\mathscr{B}$.

Fix $\sigma=\langle b_0,\dots,b_n\rangle\in\Sigma$, and let $F=\{k\le n:b_k=1\}$. Define a point $x\in X$ as follows: for each $k\in\Bbb N$,

$$x_k=\begin{cases} 1,&\text{if }k\in F\\ 0,&\text{otherwise}\;; \end{cases}$$

then $s(x)=F$, so $x\in A\cap B(\sigma)$, as desired.

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I´m really stuck, The idea could be: we take a basic in X, wich is of the form B=U1×U2×...×Un×∏Dk. – Ali Oct 26 '12 at 15:38
I´m really stuck, The idea could be: we take a basic in X, wich is of the form B=U1×U2×...×Un×∏Dk, so I need to prove that B⋂A ≠∅ if we take an element of A,lets say y, we know that a finite coordenates are equal to 1, and any basic in countable entries is the whole space Dk, so y is also in B and A is dense – Ali Oct 26 '12 at 15:51
@Ali: Yes, that’s the right idea, though it could be stated more clearly. Now that you have it, I’ll add it to my answer. – Brian M. Scott Oct 26 '12 at 15:54