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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?

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  • $\begingroup$ 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. $\endgroup$ – spaceisdarkgreen Sep 23 '17 at 4:04
  • $\begingroup$ The cardinality of each equivalence class is countable, which means the number of equivalence classes is uncountable (it has cardinality equal to the continuum). $\endgroup$ – Erick Wong Sep 23 '17 at 4:11
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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.

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