Prove that an uncountable set of must have a limit point.
If a set and is uncountable, then . Now a point is a limit point of if for every , .
Suppose for contradiction that did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?
I did not try an answer before because I was unsure of what you are working with,
But here are some observations.
If has no limits points, then every point of is not a limit point of .
If there is an open interval such that .
Because between any two numbers there is a rational number, .
Name . Note that contains exactly one point of .
How many could there be? (This is a modification of a proof by John Kelly.)
Break the real line into countably many disjoint, bounded intervals ..., [-2,1), [-1,0), [0, 1), [1, 2), ...
At least one of these intervals must contain infinitely many members of S-- otherwise S would be the union of countably many finite sets and would be countable. So...