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Let F be a countable field and B an infinite set. Let $(F^B)_0$ be the set of all functions with finite support from F to B. Is it true that $|(F^B)_0|=|B|$?

Asaf Karagila
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odnerpmocon
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1 Answers1

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Assuming the axiom of choice, yes.

To see this note that there is an injection from $B$ into $(F^B)_0$ in the obvious way; in the other direction $(F^B)_0$ is a subset (or rather, can be identified with a subset) of $(B\times F)^{<\omega}$, the set of all finite sequences from $B\times F$.

The cardinality of $(B\times F)^{<\omega}$ is $\sum_{n\in\Bbb N}|B\times F|=|B|\cdot|F|\cdot\aleph_0$, but since $B$ is infinite, you get that this just equals $|B|$.

Asaf Karagila
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  • Asaf , I think that you use $|B\times F|^n=|B\times F|$. –  Dec 06 '14 at 23:50
  • Yes, that is correct. – Asaf Karagila Dec 07 '14 at 05:09
  • @AsafKaragila doesn't this prove that for $X,Y$ with $|X|\geq |Y|\geq 2$ and $X$ infinite, $|(Y^X)_0|=|X|$? –  Nov 17 '15 at 15:01
  • @Exterior: What if $X=\Bbb N$ and $Y=\Bbb R$? – Asaf Karagila Nov 17 '15 at 15:09
  • @AsafKaragila I see. Could you explain the cardinality argument in this question? –  Nov 17 '15 at 15:10
  • @Exterior: Try http://math.stackexchange.com/questions/58548/why-are-vector-spaces-not-isomorphic-to-their-duals for starters. – Asaf Karagila Nov 17 '15 at 15:15
  • @AsafKaragila so we generally have $|M^\ast|=|k|^{\dim V}$, and in the infinite dimensional case we have $|V|=\max (|k|,\dim V)$. I am still confused at why we may conclude $\operatorname{card}V <\operatorname{card}V^\ast$ if we merely assume $\dim V \geq |k|$. If the assumption is actually a strict inequality, then I don't understand how knowing every field has a countable subfield helps. –  Nov 17 '15 at 21:54
  • @Exterior: It seems to me that you are asking a question which deserves a proper answer, and not in the comments. And since I'm going to bed, I will now leave you in the capable hands of the rest of the site. But you should write a question where you give a correct context (because frankly, I'm still not entirely sure what you're trying to understand exactly), and explain what step you seem to be missing out on. And if you can do that, you'll surely receive a good answer from one person or another. – Asaf Karagila Nov 17 '15 at 21:58
  • @AsafKaragila thanks. I asked my question here. –  Nov 17 '15 at 22:09