Is the set $T$ of all sequences of rational numbers which converge to the number $3$ countable or uncountable? I have been trying to solve the following problem:

How many rational sequences exist for $\lim_{n\to 3}$

My intuition: 
I believe the set $T$ is countable. It can be shown (in more ways than one) that $\mathbb{Q}$ is a countable set. The set $T$ is a subset of $\mathbb{Q}$ and must therefore be countable.
Is my intuition correct? 
 A: $T$ has the continuum cardinality $2^{\aleph_0}$ (Note that $2^{\aleph_0}>\aleph_0$ thus T is not countable).
Let us fix $(x_n)$ such that converges to $3$. Then, for every subsequence you take from $(x_n)$, converges again to $3$.
Thus if $A=\{B \subset \mathbb{N}: |B|=\aleph_0 \}$ you can define a subsequence $(x_{n_k})$ where $n_k$ is an enumeration of $B \in A$ for every $B \in A$.
It is well known that $|A|=2^{\aleph_0}$ and thus for Cantor-Bernstein you can conclude $|T| \geq |A|=2^{\aleph_0}$.
To conclude just observe that $T \subset \mathscr{P}(\mathbb{Q})$ and then $|T|=2^{\aleph_0}$.
A: $T$ is a set of sequences of rationals, not a subset of rationals.
The number of such sequences is actually uncountable. Clearly $3$ is unimportant; let's just pick $0$. It's well known that the number of infinite binary sequences is uncountable, so let's say we have some infinite sequence $10110101...$ that starts with $1$. 
Then the sequence
$$\frac{1}{1}, \frac{1}{10}, \frac{1}{101}, \frac{1}{1011}, \frac{1}{10110} \cdots$$
converges to $0$. We can clearly pick uncountably many such sequences, and so $T$ is uncountable.
A: I'd like to give a simple diagonalization argument that proves the statement directly.
Let's assume $T$ is countable, it's obviously infinite. So we can index all sequences $(x_n)$ in $T$ by an index $k$: $(x_n)_k$. Let's assume $n,k$ run through the positive integers, so start from $1$:
Now let's create a new sequence $(d_k)$ and define for all positive integers $k$:
$$
d_k = \begin{cases}
3,        & \text{ if }((x)_k)_k \neq 3, \\
3-\frac1k & \text{ if }((x)_k)_k = 3.
\end{cases}
$$
For any $k \ge 1$, $(d_n)$ is not the $k$-th sequence $(x_n)_k$ in $T$, because $d_k$ (the $k$-th element of $(d_n)$) was explicitely defined above to be different from the $k$-th element of $(x_n)_k$, which is $(x_k)_k$.
But obviously each $d_n$ is rational and $\lim_{n \to \infty} d_n=3$, so $(d_n)$ must be in T.
This is a contradiction and proves that $T$ is actually not countable.
