this literally ruins everything
take (0, 1) upper bound 1, lower bound 0.
consider n \in (0, 1)
given n != 0 or 1, for all n otherwise 50% of all real numbers in (0,1) are both greater than and less than n, since 1, 0 \notin (0,1) then this is true all the time for all n.
proof:
f(n) = n
g(n) = n + 1
f(n) < g(n)
lim(n->inf) : inf < inf
given x = lim r ->inf
lim(f(n), g(n)) = x
x < x
contradiction
thus, lim f = lim g
given that there are infinite reals between (0,1)
for any n, as a point on (0,1) can signify a cut where:
M: (0, a) and N: (a, 1) are both infinite and therefore can map onto each other.
M->N has a one to one correspondence.
and it's 1/2 instead of 1/3 because of the earlier result (which should be obvious)