Cardinality of set of sequences from $\mathbb{Q}$ that converge to $0$ I'm looking for cardinality of a set of sequences from $\mathbb{Q}$ which are convergent to $0$. 
I think the answer is continuum, but I don't know how to prove it.
 A: There are certainly at most $c$ of them, since there are only $\aleph_0^{\aleph_0}=c$ sequences in $\Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
A: There are (at least) as many such sequences as there are subsets of $\Bbb N$. Indeed, given $S\subseteq \Bbb N$, we can let
$$a_n=\begin{cases}\frac1n&n\in S\\-\frac 1n&n\notin S\end{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|\Bbb Q|=|\Bbb N|$). By mapping $(a_n)_{n\in\Bbb N}$ to $(a_1+\ldots +a_n)_{n\in \Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $\Bbb N$, of which there are at most continuum-many.
A: Consider the set $L$ of sequences $(x_n)_{n\in\mathbb{N}}$, with $x_n \in \{0, \frac{1}{n}\}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $\mathbb{Q}^{\mathbb{N}}$.
Then $|L| = |2^{\mathbb{N}}| = |\mathbb{R}| \leq |M| \leq |\mathbb{Q}^{\mathbb{N}}| \leq |(2^{\mathbb{N}})^{\mathbb{N}}| = |2^{\mathbb{N}^2}| = |2^{\mathbb{N}}| = |\mathbb{R}|$, so $|M| = |\mathbb{R}|$.
A: It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
