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Math Help - cardinality

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by sfitz View Post
    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 |A| \leq |C| because we know there exists injective maps \mu: A\mapsto B and \eta: B\mapsto C so \eta \circ \mu : A\mapsto C is injective. Which means |A|\leq |C|. Now you argue that it is not possible to have |A| = |C|. Assume that if \phi: A\mapsto C is bijection. Then  \eta^{-1} \circ \phi :A\mapsto B is a bijection contradicting the fact that |A| < |B|.


    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. F is a set of all function from K to where?
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  3. #3
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    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!
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