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?

>>21
I'm the person who wrote >>16.
Further, I can't use the fact that S has, to quote and emphasize >>16, "AT LEAST two elements" in the case that it has INFINITELY MANY elements? Are you fucking retarded? If it's infinite, we can choose two elements and call them a and b. We can then use these two elements to describe yi as follows:
If xii = a, yi = b
else yi = a.

Christ.