Given: S is a nondenumerable set (uncountable)
T is a denumerable subset of S
Prove: (S - T) ~ S. (S - T is nondenumerable)
Well, no, precisely because there are many levels.Because they are both nondenumerable, we can still say (S - T) ~ S, correct?
To show that S - T ~ S, you can consider a denumerable subset T0 of S - T and construct a bijection between S - T and S as follows. On S - T - T0 the bijection is identity, and on T0 it is a bijection between T0 and (T0 union T).