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Suppose $a,b$ are cardinals where $a$ is finite and $b$ is infinite. I want to prove that $b^a=b$. The book gives a hint saying to use repeated multiplication of cardinals to do it.

I have proved the result using induction and the fact that $b^{a+1}=b^ab^1$ but I am just noticing that we have not proved this yet, in fact its the next question. So is there some way to prove my question without this property?

So I am trying to find the cardinal number of the set of all functions from $A$ to $B$, where $b=|B|, a=|A|.$

H_B
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1 Answers1

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You need two facts for this proof, one of which is fairly easy to prove by hand.

  1. There is a bijection between ${}^AB\times B$ and ${}^{A\cup\{a\}}B$, where $a\notin A$. From this we can conclude that $b^a\cdot b=b^{a+1}$.

    This is really just writing down the simplest, most obvious bijection and showing that it is a bijection, which is not very hard.

  2. If $X$ is an infinite set, then $X\times X$ and $X$ have the same cardinality. Namely, $a=a^2$ for every infinite $a$.

    This is a bit trickier, since this statement is in fact equivalent to the axiom of choice, so we can't quite prove it out of the blue in this sort of generality; although we can prove it for some sets, like $\Bbb N$ or $\Bbb R$ for example.

Asaf Karagila
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  • Ok this is exactly what I did the first time, but this is question #2 in the book, and question #3 is prove that, for cardinalities, $a^ba^c=a^{b+c}$, so I thought there might be another somewhat "basic" way. Thank you again. – H_B Dec 01 '14 at 20:36
  • Yes, that is an even more general claim, whose proof is essentially the same. – Asaf Karagila Dec 01 '14 at 20:52
  • Also, we just proved that for infinite cardinalities $a*a=a$ :)

    one quick question, does ${}^AB\times B =B^A\times B$? I have never seen this notation.

     – H_B Dec 01 '14 at 20:54
  • Oh, it's an alternative notation for the set of functions from $A$ to $B$; we're just using it in the course I'm teaching, so I chose to use it here as well for some reason. – Asaf Karagila Dec 01 '14 at 20:55
  • So then $B^A\times B = B^A\times B^1=B^A\times B^{{a}}$

    So we want a bijection from $B^A\times B^{{a}}$ to $B^{A\cup {a}}$

     – H_B Dec 01 '14 at 21:08
  • Yes. Try first to see why $B^2$ and $B\times B$ have a bijection between them. From there it's a walk in the park. – Asaf Karagila Dec 01 '14 at 21:11
  • So for my bijection I am sending functions from $B^A\times B^{{a}}$ to functions in $B^{A\cup {a}}$? – H_B Dec 01 '14 at 21:11
  • Note that $B^A\times B^{{a}}$ is a set of ordered pairs of functions, whereas $B^{A\cup{a}}$ is a set of functions. – Asaf Karagila Dec 01 '14 at 21:12
  • I am thinking the char function will do the job

    $g:B^{A\cup {a}}\rightarrow B^A\times B^{{a}}$,$g(f)=(\pi_A f,\pi_{{a}} f)$

     – H_B Dec 01 '14 at 21:27
  • Assuming $\pi_X$ is the restriction of $f$ to $X$, you're right. Yes. – Asaf Karagila Dec 01 '14 at 21:34
  • Wow cool. Thanks again. Remember if you get sick of things, we need more set theorist in Portland;) – H_B Dec 01 '14 at 21:38