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Suppose $m, n$ are infinite ordinal numbers. $$a) m=n → |m|=|n|$$ $$b)|m|=|n| →m=n$$ $$c)m<n→ |m|<|n|$$ $$d)|\max{(m,n)}|< |m|+|n|$$ $$e)|m|<|n| →|m|^{|n|}<|n|^m$$

Which of the above statements are true? (a) looks true but I do not know the way to work it out. Please help

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What is your definition of cardinal? –  Brian M. Scott Jul 1 '13 at 11:14
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That doesn’t help. If $m$ is a cardinal, what’s the difference between $m$ and $|m|$? –  Brian M. Scott Jul 1 '13 at 11:16
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There’s no way to answer the question unless you can find out just what definition of cardinal number the person who composed that exam paper expected you to use. If nothing more was said on the paper, the person who composed it must have expected students to know a particular definition. –  Brian M. Scott Jul 1 '13 at 11:21
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No one is sending out exams to arbitrary people. If you are looking at this exam then you should have studied some course, some material, something. And in that something you should find the definition of a cardinal number which will then tell you what the answer should be. –  Asaf Karagila Jul 1 '13 at 11:22
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Before I'm getting into writing n answer. Do you know what an ordinal number is? Do you know what is a successor ordinal, and what is a limit ordinal? –  Asaf Karagila Jul 1 '13 at 11:41

1 Answer 1

up vote 1 down vote accepted

Hints:

  1. Recall that $|x|=|y|$ is an equivalence relation.
  2. Recall that $|\Bbb N|=|\Bbb N\setminus\{0\}|$, so adding or removing one element does not change the cardinality. You said that you know what is a successor ordinal, this should give you a counterexample.
  3. The counterexample to the previous statement should give you a counterexample to this one as well.
  4. Recall that addition of cardinals satisfy $|a|+|b|=\max\{|a|,|b|\}$.
  5. Consider $\omega$ and $\frak c$ (as suggested by Chris Eagle in the comments).
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For (e), just take $\aleph_0<2^{\aleph_0}$. –  Chris Eagle Jul 1 '13 at 12:04
    
@Chris: You're right. I was thinking about equality, rather than reversing the inequality sign. –  Asaf Karagila Jul 1 '13 at 12:05

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