A proposition...

For all x in R, there exists a sequence a_n in Q such that the limit of a_n as n->+inf = x

This seems trivially false to me just from cardinality arguments (e.g. if you assume it's true then the cardinality of Q being less than the cardinality of R immediately implies that there are two different x in R that must have the same limit in Q), but I'll be damned if I can prove it rigorously.  Is there an easier way?

>>24
Isn't this basically the definition of the axiom of choice for laypeople?  That you can pick an item from an infinite set?

So >>16 is totally obvious that, if S contains exactly 2 elements, it works.  But given that I'm so fucking gunshy of assuming that things that are obvious are also correct (now that I've been raped by analysis), I'm worried that assuming you can proceed from 2 to N to countably infinite to uncountably infinite elements is allowed.  Assume ZFC and all that, for the sake of my battered-child approach to mathematics, how would one show that moving from |S| = 2 to |S|>2 is ok in such a tooth-grindingly pedantic manner that no TA could realistically take issue with it?