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In Lang "Real and Functional analysis" is demonstrated that given a countable set $A$ and a function $f: A \rightarrow B$ which is surjective on $B$, then $B$ is finite or countable.

Proof: Consider $y \in B$ then there exists a non void set $F_y= \{x \in A | f(x)=y \}$, consider one element of the set , say $x_y \in F_y$. Now the assosiation $y \rightarrow x_y$ is injective from $B \rightarrow A$ (the proof of injectivity is easy) then, for the definiton of countability we have that also $B$ is countable.

My questions are:

  • It seems to me that this proof uses the axiom of choice to choose the elements in the family $F_y$, is this correct?

  • If yes, there are proof of the same "theorem" without the AOC?

  • I've heard of a theorem which says that if there exist a pair of injection $i_1,i_2, \ \ i_1:A \rightarrow B \ \ i_2:B \rightarrow A$ then there is a bijection $ f:A \rightarrow B$. And that this theorem can be demonstrated without AOC, is this true? There is a similar theorem for surjective functions?

Andrés E. Caicedo
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HaroldF
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1 Answers1

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(1,2) As written, the proof certainly seems to appeal to the axiom of choice for choosing each of the $x_y$s. But this is not necessary and easily eliminated: since you know that $A$ is countable, you can choose a bijection $A\leftrightarrow \mathbb N$ once and for all, and then for every $y$ let $x_y$ be the element in $F_y$ that has the smallest index.

(3) Yes, this is the Cantor-Schröder-Bernstein theorem, and it does not depend on the axiom of choice. The corresponding theorem for surjections does require AC.

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hmakholm left over Monica
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  • Thank you! In (1,2) isn't your argumentation the Axiom of Countable choice? https://en.wikipedia.org/wiki/Axiom_of_countable_choice

    Sorry if i'm silly, but i'm new to this kind of "set theory" and I'm very interested

     – HaroldF Jan 23 '16 at 15:26
  • @HaroldF: No, the axiom of countable choice allows you to choose one from each of countably many non-empty sets (but those sets can individually be as large or small as you want). That is different from what we need here -- namely, choosing a bijection $A\leftrightarrow \mathbb N$ gives us for free a way to choose an element of each of the uncountably many nonempty subsets of $A$ (but each of those subsets happens to be finite or countable). – hmakholm left over Monica Jan 23 '16 at 15:29
  • We can't appeal to countable choice instead here, because we need to make a choice for each element $y\in B$, and at this point we don't know yet that there are only countably many $y$s. – hmakholm left over Monica Jan 23 '16 at 15:30
  • Oh sorry, probably I expressed my self roughly. I mean, the fact that we can choose the lowest element in a subset of $\mathbb{N}$ for a family of subsets of $\mathbb{N}$ isn't again the AOC, or the AOCountableC? – HaroldF Jan 23 '16 at 15:34
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    @HaroldF: No, it's a basic property of $\mathbb N$ that every subset of it is either empty or has a (unique) smallest element. We can prove this by mathematical induction: Assume that $A\subseteq \mathbb N$ has no smallest element; then by induction on $n$ we have that $m\notin A$ for all $m\le n$ and in particular $n\notin A$. Therefore $A=\varnothing$. – hmakholm left over Monica Jan 23 '16 at 15:37
  • Oh, ok now i see. Thank you again! – HaroldF Jan 23 '16 at 15:43