How many equivalence classes are there in this closeness relation on sequences?

Question

I had been watching a pop-math video and I noticed that they made the following assumption.

Let $\Omega$ be the set of sequences on $\{ 0, 1\}$. Call two sequences $t_1, t_2 \in \Omega$ close if they differ at finitely many places.

Then, there are only countably many equivalence classes of the closeness relation.

How do I show this?

Answer 1

There are uncountably many equivalence classes. Observe that there are only a countable number of elements in each equivalence class, so if there were a countable number of equivalance classes then a countable union of countable sets would be uncountable.