Subsequence equivalence: $(a_n)$ and $(b_n)$ in $\{0,1\}^\mathbb N$ are equivalent if they have some subsequences which are equal

Question

Define two sequences $(a_n), (b_n) \in \{0, 1\}^\mathbb{N}$ are equivalent iff they have some subsequences that are equal.

How many equivalence classes are there? Is it uncountable?

It might be countable (unlikely but possible), since each equivalence class is uncountable.

Motivation: I wanted to count the number of "essentially different" ways for a rational sequence to converge to $0$. Two convergent sequences are equivalent iff they have some subsequences that are equal. I could determine that there are uncountably many of these by this construction:

$$a_n\neq b_n \in \left(\frac{1}{2^{n+1}}, \frac{1}{2^{n}}\right]\cap\mathbb{Q}$$ then any two such constructed $(a_n)$, $(b_n)$ will be inequivalent but convergent to $0$. There are clearly $\mathbb{N}^\mathbb{N}$ of them.

Answer 1

There are two things here:

1. The relation is not even transitive. Consider the constant sequence $1$, the constant sequence $0$, and any alternating sequence $(a_n)$. Then $(a_n)$ is equivalent to both $(1)$ and $(0)$, but they are not equivalent to one another.

2. You are forgetting that the sequences of rationals can obtain many more values than just $\{0,1\}$.

But now to your actual question, note that there is a family $\{A_r\subseteq\Bbb N\mid r\in\Bbb R\}$ such that for $r\neq t$, the intersection $A_r\cap A_t$ is finite. Use this family to find $2^{\aleph_0}$ essentially different subsequences of $\frac1n$ which converge to $0$.

Answer 2

This isn't an equivalence relation, because it is not transitive. For instance, the sequence $a_n=0$ is "equivalent" to the sequence $b_n=\frac{1+(-1)^n}2$, which is in turn "equivalent" to $c_n=1$. Yet, $a$ and $c$ are not "equivalent".