# Obvoiously True (Countability)

1. Oct 20, 2009

### CoachZ

I'm trying to show that any uncountable set has a countable subset.
First, let me point out that the distinction here between at most countable and countable is applied in this instance. At most countable implies either finite or countable, and countable is obvious.

Starting off, let X = the uncountable set, and choose some subset A of X, s.t. X\A = B. Not really sure where to go from this point. I had an idea that this could be proved inductively, but I'm not exactly sure how the process goes... Any suggestions?

2. Oct 20, 2009

### VeeEight

I'm not really sure where you're proof is going, but can you not just select elements from X to form an countable subset? For example, since X is uncountable, it is nonempty and thus you can find some a1 in X. Similarily, X \ {a1} is also uncountable and hence nonempty, so you can find some a2 in X \ {a1}. Each subset of the form X \ {a1, a2, ...} is uncountable and hence nonempty, so you can always find some element ai in it.