Thread: cardinality

I have two problems:

1. Let A, B, and C be sets. Suppose that card A < card B and card B < card C. Prove that card A < card C.

Can I do this by contradiction?:
Assume card A < card B, card B < card C, and by contradiction, that card A > C.
We are give that card A < card B.
By transitivity of the relation <, card B < card C and card C < card A implies that card B < card A.
Contradiction, because card A < card B and card B < card A can't both be true.

Is this right or am I missing something?

2. Let K be any set, and let F be the set of all functions with domain K. Then card K < card F.

I don't know how to start with this one. It makes sense, because there are infinitely many functions for each possibility of the codomain... But I don't know how to say that. Any hints?

Thank you!

Originally Posted by sfitz

1. Let A, B, and C be sets. Suppose that card A < card B and card B < card C. Prove that card A < card C.

The first step is to note that because we know there exists injective maps and so is injective. Which means . Now you argue that it is not possible to have . Assume that if is bijection. Then is a bijection contradicting the fact that .

2. Let K be any set, and let F be the set of all functions with domain K. Then card K < card F.

This is not a good set. is a set of all function from to where?

The first one makes sense, thank you! What is the problem with what I did? Or does it make sense, but just not as good a proof?

For the second one, that's all the problem said. I think from K to anything, all the different possibilities. There's no other information.

Thank you!