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?

>>8
It's not a countable union.

Can use the diagonal argument if you wanted

consider a bijection f from the naturals to sequences of rationals....

f(1)=a11,a12,a13.....
f(2)=a21,a22,a23.....
.
.
.
f(i)=ai1,ai2,ai3....
.
.

Now construct a sequence of rationals call this S

s_i = a_ii + 1 mod 10

There doesn't exist any natural number n s.t f(n)=S as s_n !=a_nn

Diagonal argument basically.