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I try to prove the following exercise without using the axiom of choice. On the second item, the proof that I found in the books, to define the function $g:Y\to X$, for every $y\in Y$, due to the surjective of $f$ they choose $x=g(y)\in X$ such that $f(x)=y$.

Let $f:X\to Y$, prove.

  1. If $Y$ is finite and $f$ injective, then $X$ is finite.
  2. If $X$ is finite and $f$ surjective, then $Y$ is finite.

Proof:

  1. Done.

  2. Let $X=\{x_1,\ldots,x_n\}$. Given any $y\in Y$, because $f$ is surjective, then exists $k\in I_n=\{1,\ldots,n\}$ such that $y=f(x_k)$. Consider the set $$A_y:=\{i\in I_n:f(x_i)=y\}$$

    Then $A_y\neq\emptyset$ because $k\in A$, so, by the Well-ordering principle, $A_y$ has a minimal element, let this element $i_y$, then $$f(x_{i_y})=y.$$

    Define $g:Y\to X$ by $g(y):=x_{i_y}$. Im particular we get $$f(g(y))=f(x_{i_y})=y,\quad \forall y\in Y.$$

Prove that $g$ is injective. Let $y,z\in Y$ such that $g(y)=g(z)$, then $$g(y)=g(z)\Longrightarrow f(g(y))=f(g(z)) \Longrightarrow y=z.$$

Therefore, $g$ is injective, then by the first item, $Y$ is finite.

Is it right?

Andrés E. Caicedo
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Framate
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3 Answers3

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Yes, your proof is fine. Although it is in fact a bit complicated.

If you have proved that if $X$ is finite, then $\mathcal P(X)$ is finite, then you can just appeal to this, as $y\mapsto f^{-1}(y)$ is an injection from $Y$ into $\mathcal P(X)$.

Asaf Karagila
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  • Thank you, that is a very nice and simple answer. But acording to the material covered, it has not proven yet that if $X$ is finte then $P(X)$ is finite(it is an exercise at the end of the section), having done this, i can use this as a second solution – Framate Feb 25 '20 at 11:23
  • Yes, note that the fact about power sets is entirely independent of the statement you're trying to prove here. So there will be no circularity. – Asaf Karagila Feb 25 '20 at 11:25
  • Note that the OP is essentially proving that every if $X$ is any finite set then any partition $\rho$ of $X$ if finite (only a finite number of blocks). And of course $\rho \subset \mathcal P(X)$. Just saying... – CopyPasteIt Feb 25 '20 at 17:22
  • @CopyPasteIt: Yes, this is the alternative proof I've suggested. You're right that this is equivalent. Nevertheless, partitions tend to be significantly smaller compared to the power set (e.g. any partition of a countable set is countable). – Asaf Karagila Feb 25 '20 at 18:33
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Your proof is correct. The reason you do not need choice is because you have a well-order on $X$ (which exists without choice because $X$ is finite). So you can do essentially the same argument as at the top of your post, "...choose $x$ such that $f(x) = y$...", only now "choose" does not invoke choice because of the well-order on $X$.

In fact, your proof generalises to the following statement. If $X$ has a well-order and $f: X \to Y$ is surjective, then there is a section $g: Y \to X$ (i.e. injective $g$, s.t. $fg$ is the identity).

Mark Kamsma
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You do it very complicated. Suppose $|Y|=|X|+1$. Since it's surjective, if $y\in Y$, there is $x\in X$ s.t. $y=f(x)$. But then, $|X|$ element will have $|X|+1$ images, which is impossible (an element can not have several images). Therefore $|Y|\leq |X|$.

Surb
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  • While not essentially wrong, there is a case to argue this answer is misleading without additional information. Some of the times we prove that the cardinals behave the way they should by arguing first about injections and surjections and all of that. So making the appeal to size consideration is now circular. Without knowing the exact context in which the question is framed, it's hard to say, though. – Asaf Karagila Feb 25 '20 at 11:08
  • We can wait until the OP clarifies... – Asaf Karagila Feb 25 '20 at 11:11
  • @AsafKaragila thank you, the context is a course on introduction to real analysis, finite and infinite sets, i am no using cardinals numbers, just cardinal as the number of elements of a set. – Framate Feb 25 '20 at 11:17