What is the cardinality of $A$?

Question

Consider a set $$A = \left \{(x_n)_{n \geq 1}\ |\ x_n \in \{0,1 \}\ \text {for all}\ n \geq 1\ \text {and}\ x_n = 0\ \text {for all but finitely many}\ n \in \Bbb N \right \}.$$

Then what is the cardinality of $A$?

Clearly it doesn't exceed $2^{\aleph_0}.$

Answer 1

HINT: Each member of $A$ is the characteristic (or indicator) function of a finite subset of $\Bbb Z^+$.

Answer 2

For each member of $A$, it is tempting to assign the following nonnegative integer: $$\sum_{n=0}^\infty x_{n+1} 2^n$$ i.e. treating the the zeros and ones as binary digits of a whole number. The series is essentially a finite sum because almost all $x_n$ are zero.

This gives an explicit bijection (prove it) between $A$ and $\mathbb{Z}^{\ge 0}$, so by definition the cardinality of $A$ is only $\aleph_0$.