# A simple question about counting and sets

1. Aug 3, 2006

### siifuthun

This is the question, and we're supposed to answer if it's true or false:

If A is a countably infinite set, and A is a proper subset of another set B,
then B is uncountable.

I thought this was false, because if A is infinite and countable, then B should also be infinite and countable in the same way A is if it's a proper subset of B. Could we list elements of A, then elements of B that are not contained in A?

2. Aug 3, 2006

### Data

Which of $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}$ are contained in which? Which ones are countable?

3. Aug 3, 2006

### siifuthun

If A is the set containing natural numbers, and B is the set containing real numbers, then B would have to be uncountable right?

4. Aug 3, 2006

### d_leet

But what if A is the natural numbers and B is the integers?

5. Aug 3, 2006

### siifuthun

That's a good point. In that case it would be countable. So I guess in that case the answer is false, seeing that B can be countable or uncountable.

Thanks.

6. Aug 3, 2006

### StatusX

The defining property of infinite sets is that they have proper subsets of the same cardinality as the set.

7. Aug 3, 2006

### siifuthun

So, since B is infinite (and it should be if A is, correct?), then it should have the same cardinality as A - and we know the cardinality of A because it's countable, meaning that B would be countable in that case?

8. Aug 3, 2006

### StatusX

Not necessarily. It just means that for any infinite set B, there exists a proper subset A of B with the same cardinality as B. Thus if B is countably infinite, it has a countably infinite proper subset A, and so these form a counterexample to the statement in your first post. It doesn't mean that all sets containing A are countably infinite, as for example, the reals contain the integers, but are not countable.