# Thread: Nondenumerable - Denumerable = Nondenumerable

1. ## Nondenumerable - Denumerable = Nondenumerable

Given: S is a nondenumerable set (uncountable)
T is a denumerable subset of S

Prove: (S - T) ~ S. (S - T is nondenumerable)

2. Is this true $T\cup (S\setminus T)=S?$

What if $(S\setminus T)$ were denumerable?

3. isn't that just another way of wording it?
it would be a denumerable subset of a nondenumerable set, which i believe is acceptable...just not the other way around

4. Originally Posted by DavyHilbert
isn't that just another way of wording it?
it would be a denumerable subset of a nondenumerable set, which i believe is acceptable...just not the other way around
No it in not a reword.
The union of two denumerable sets is a denumerable set.
But $S$ is not denumerable.

5. ahhh, thanks. understood.

Because they are both nondenumerable, we can still say (S - T) ~ S, correct? Even though there are different "levels" of nondenumerable?

6. Because they are both nondenumerable, we can still say (S - T) ~ S, correct?
Well, no, precisely because there are many levels.

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).