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

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?

• Not clear when they made that assumption (although I didn't watch the whole video, just that segment) and I don't think it's true. – spaceisdarkgreen Sep 23 '17 at 4:04
• The cardinality of each equivalence class is countable, which means the number of equivalence classes is uncountable (it has cardinality equal to the continuum). – Erick Wong Sep 23 '17 at 4:11