TITLE: cardinality of Y=$\{(x_n)\ : x_n=1\}$ for at most finitely many $n$
QUESTION [5 upvotes]: Let $X$ be the set of all binary sequences i.e. $X=\{(x_n) : x_n \in \{0,1\}, n \in \mathbb{N}\}$ and Y be its subset such that  Y=$\{(x_n)\in X : x_n=1 \text{ for at most finitely many } n \}$. It is clear that $X$ is uncountable using Cantor's diagonalization argument. 
But Y must be countable right? As I think we can enumerate it and thus have a 1-1 function with $\mathbb{N}$. Am I right?

REPLY [2 votes]: Recall that the countable union of countable sets is again countable.  If we define $U_k = \{(x_n) | x_n = 0\; \forall k > N\}$, then we can observe the following:
1) Each $U_k$ is a subset of $Y$ since each $(x_n)\in U_k$ contains at most $k$ entries which are equal to 1.  This means that $\bigcup_{k\in\mathbb{N}} U_k \subseteq Y$.
2) Each $U_k$ is finite.  In fact, $|U_k| = 2^k$.
3) $Y \subseteq \bigcup_{k\in \mathbb{N}} U_k$.
1 and 3 together show that $Y = \bigcup_{k\in \mathbb{N}} U_k$.  Altogether we have shown that $Y$ is the countable union of countable sets, and hence $Y$ itself is countable.