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Math Help - Cardinals and order problems

  1. #1
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    Cardinals and order problems

    I have problems with cardinals and their order.
    Problem:
    Let K,L,M be sets so that card(K) = k, card(L) = l and card(M) = m.
    Show that from the condition k < l does not follow, that
    a) k + m < l + m
    b) k*m < l*m
    c) m^k < m^l

    (Notice: the symbol is really "smaller than" not "smaller or as great as".)

    I have tried to solve this for hours, and I really don't get a grip of it. Can anyone even help me to right direction?
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  2. #2
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    a) k + m < l + m

    Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.

    b) k*m < l*m

    Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.

    c) m^k < m^l

    Hint: Let m = 0. Then find suitable k and l to contradict the above.

    Or another approach:

    Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.
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  3. #3
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    Okey, it didn't even come to my mind, that I can assume some of the cardinals to be infinite.

    I was thinking to proceed like this: "Let k be finite and l be infinite...", but I still don't manage to solve the problems.
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  4. #4
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    Quote Originally Posted by Ester View Post
    Let k be finite and l be infinite
    Not quite. Instead, think about this question (your course or book should have covered this by the time you got to this excercise): What happens when you add a finite cardinal to an infinite cardinal?
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  5. #5
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    Quote Originally Posted by MoeBlee View Post
    What happens when you add a finite cardinal to an infinite cardinal?
    Okey.. Let k be finite cardinal and l infinite cardinal.
    I think that happens something like: k + l = l. Am I correct?
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  6. #6
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    Quote Originally Posted by Ester View Post
    Okey.. Let k be finite cardinal and l infinite cardinal.
    I think that happens something like: k + l = l. Am I correct?
    Yes, in the sense that a finite cardinal added to an infinite cardinal is just the infinite cardinal. But in this case, it's not l that you want to be infinite. Now just determine which of k, l, and m should be finite and infinite and you're almost done.
    Last edited by MoeBlee; April 12th 2010 at 07:15 AM.
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  7. #7
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    Quote Originally Posted by MoeBlee View Post
    Now just determine which of k, l, and m should be finite and infinite and you're almost done.
    I have this condition: k < l, so in my mind, if I assume that l is finite cardinal, then k also have to be finite, because k < l < N (aleph). So the m should be infinite..?
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  8. #8
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    Right. Let k < l both finite and m infinite.
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  9. #9
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    Is it really this easy: Let k,l be finite, and m infinite. Now k + m = m = l + m.
    And so it can't be k + m < l + m.

    Now I think I have to prove that there is bijection between k + m and m, because we don't have this kind of theory (finite + infinite = infinite) in our course.
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  10. #10
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    Yes, if you also say that you've chosen k < l.

    So I would write it this way:

    Let k and l both be finite with k < l. Let m be infinite. Then k+m = l+m, so it is not the case that k+m < l+m. So it is not the case that, for all cardinals k, l, and m, we have k < l implies k+m < l+m.

    But remember, this depends on having already proven (in your course or in your textbook) that a finite k added to an infinite m is just m.
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