Is this true
What if were denumerable?
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).