## Infinite cardinal numbers.

Prove that if is an infinite cardinal number, then and .

**Solution:** Let . Then .

We denote the set containing non-void subsets of . For any non-void set we denote .

Let . The set , being infinite, has a infinite countable subset; so there exists injective. Then the set is countable because is bijective and . Furthermore we have . Therefore we have a function bijective. This shows that the set if non-void.

Consider the binary relation

( The function is an extension of if and )

We prove that (\mathcal F,\mathcal R) $ is a partially ordered set (i.e. order relation).

: is one of its extensions because and .

: We have and . Because we must have and , so and . Therefore .

: We have . Furthermore, and and . Take . We have . So . This means that is an extension of . Therefore .

We have proved that is an order on .

Take a totally ordered set. It’s obvious that for any The function is bijective. Even more, for the function is an extension for or is an extension for .

From above, we get that if , then the function defined for any by , for all with , is bijective.

is a bijective extension for all . Therefore is an upper bound for . Since was chosen randomly, we proved that any totally ordered set has an upper bound.

Using **Zorn**‘s lemma, we find that has at least one maximal element . This means that is non-void and cannot be extended

We’ll show that is finite. Suppose that is infinite. then we can find an injection . Therefore, and are countably infinite, so we can find a bijection . Let defined for all by

.

Then is in and extends . This contradicts the maximality of .

Therefore, is finite and is infinite. From here we get the equality

. Therefore , where .