Clarification wanted: Let $T$ be the set of all infinite sequences of $0$'s and $1$'s with finitely many $1$'s. Prove that $T$ is denumerable. I think I'm misunderstanding the following proposition.

Let $T$ be the set of all infinite sequence of $0$'s and $1$'s with only finitely many $1$'s. Prove that $T$ is denumerable. 

I'm also given the following lemma to use:

Let $\{A_{i}| i\in\mathbb{N}\}$ be a denumerable family of finite non-nonempty sets which are pairwise disjoint, then $\bigcup _{i\in\mathbb{N}}A_{i}$ is denumberable. 


I'm not quite seeing how adding the condition that there are finitely many $1$'s makes this the set $T$ countable, since there is still a possibility of infintely many $0$'s. It still looks susceptible to the diagonalization argument. 
Could anyone clear up why the the set is countable? 

Use the fact that "a countable union of countable sets is countable," several times.


*

*The set of binary strings with finitely many 1s is the union of binary strings with 1 1, 2 1s, 3 1s, 4s, etc.

*The set of binary strings with $n$ 1s is the union of binary strings with $n$ 1s where the greatest 1 is in the $k$ position, the $k+1$ position, the $k+2$ position, etc.

*The set of binary strings with $n$ 1s before the $k$ position and the rest all $0$s is quite clearly less than $2^{k+1}$. Congratulations, you are done.


On further thought you can skip step (1) entirely, but I thought I would include it since it really was the first simplification I thought to make, and you should ruthlessly simplify using the pretty powerful theorem that countable union of countable sets is countable.
