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We are given $A$ which is an infinite set, and $F = \{f : A \to \{0, 1\} \}$. I need to prove there exists a surjective function $g : F \to A$.

My solution was this:

Define $g(f) = \omega$ and $f \in F$ such that $f (\omega) = 1$ for some $\omega \in A$. Now for each $\alpha \in A$, there exists $h : A \to \{0, 1\}$ such that only $h(\alpha) = 1$ and $0$ otherwise. This implies $g (h) = \alpha$. And thus $g$ is surjective.

Is my proof valid, since $g (f)$ is not uniquely defined in my example and can be any element $\omega$ of $A$ as long as $f (\omega) = 1$? I thought of making $\omega$ the smallest element such that $f (\omega) = 1$, but if $A$ is an uncountable infinite set, I think a smallest element cannot be defined. It would also be very helpful if someone can share any other proofs.

Frederik vom Ende
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Shaurya Agarwal
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  • The set $F$ is essentially the power set of $A$ minus two elements (the empty set and the whole set). It is easy to construct a surjection from $P(A)$ to $A$ that does not depend on the images of the empty set and of $A$. – Arturo Magidin Aug 02 '24 at 18:54
  • @ArturoMagidin Why does $F$ not contain $\emptyset$ and $A$? – K. Jiang Aug 02 '24 at 18:58
  • "but if $A$ is an uncountable infinite set, I think a smallest element cannot be defined" Note that even if $A$ were countable, there need not be a "smallest element". The idea of "smallest element" implies there exists some order on the set $A$, which there need not be. – K. Jiang Aug 02 '24 at 19:07
  • @K.Jiang I thought $F$ only contained surjective functions onto ${0,1}$. I may have misread it. – Arturo Magidin Aug 02 '24 at 19:10
  • @ArturoMagidin There is no such constraint on $F$ so $\phi$ and $A$ are also elements in $F$ – Shaurya Agarwal Aug 02 '24 at 22:47

2 Answers2

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Indeed, your $g : F \to A$ is not well defined because it does not specify a unique element of $A$ as the image of a given $f$.

Define $g$ as follows: for $f \in F$, if $f$ maps multiple elements of $A$ to $1$ or everything to $0$, then let $g (f) = a_0$, where $a_0 \in A$ is some arbitrary element; if $f$ maps a single element of $A$, say $a$, to $1$, then let $g (f) = a$. You can check that $g$ is well defined.

To show that $g : F \to A$ is surjective, we show that every element $a \in A$ has a preimage. This is straightforward: for any $a \in A$, consider the function $f_a \in F$ that maps $a$ to $1$ and everything else to $0$. Clearly, $g (f_a) = a$, because $a$ is the only element that gets mapped to $1$.

K. Jiang
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    I see, in the case when $f$ maps multiple elements to 1, you defined $g(f)$ as some arbitrary element in $A$ whereas I defined it as some arbitrary element $\omega$ in $A$ such that $f(\omega)$ also has to be 1. But can we call your function well defined when $a_0$ is not fixed but can take any value in $A$? – Shaurya Agarwal Aug 02 '24 at 19:10
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    @ShauryaAgarwal No, you pick one $a_0 \in A$ beforehand, and all those functions are mapped to that same $a_0$ (i.e., it is fixed). We are not choosing a custom one for each function $f$ (you could with the axiom of choice, but that is not needed here). – K. Jiang Aug 02 '24 at 19:12
  • @K.Jiang, when you pick one $a_0\in A$ beforehand, are not you using the axiom of choice since $A$ is an infinite set and you are picking an arbitrary element of $A,?$ – Angelo Aug 03 '24 at 07:16
  • @Angelo You do not need choice to make $1$ arbitrary choice (or more generally, finitely many choices), even if it is from an infinite set. Choice is needed when infinitely many arbitrary choices need to be made. – K. Jiang Aug 03 '24 at 12:42
  • @K.Jiang, could you explain me why finitely many arbitrary choices from an infinite set do not need the use of the axiom of choice? What is the reason? – Angelo Aug 03 '24 at 18:16
  • @Angelo You can read detailed explanations here: https://math.stackexchange.com/a/29383 and here: https://mathoverflow.net/a/32546. – K. Jiang Aug 03 '24 at 18:24
  • @K.Jiang, I had already read detailed explanations like those ones you have recommended to me, but nobody explained me how the existence of an element of a set allows us to find it explicitly in order to get a choice function. Please, could you explain it to me in a clear and precise way? – Angelo Aug 03 '24 at 19:15
  • @Angelo This probably isn't the best place to discuss this, but to your point: there may not be an explicit element that gets selected (indeed, a proof for a "finite" axiom of choice - which does not require the axiom of choice - is not constructive). However, the mere fact that $A \ne \emptyset$ implies the existence of some element $a_0 \in A$. Select it, and you are finished. If you want to discuss further, let's move to chat: https://chat.stackexchange.com/rooms/36/mathematics. – K. Jiang Aug 03 '24 at 21:39
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I think you are on the right track but not quite there. As you said $g$ is not uniquely defined because $g(f)$ can equal 2 different elements of $A$. One way to fix this would be to ignore all the $f$ where $f(x) =1$ for more than one $x$.

Formally, pick a $\beta \in A$. If $f\in F$ such that $|f^{-1}(1)| \neq 1$, then define $g(f) =\beta$. Otherwise, $f^{-1}(1)$ is uniquely defined so we can say $g(f) = f^{-1}(1)$.

A “simpler” solution would just be to show that $|F| \ge |A|$ so that the existence of a surjective function is directly implied (If you dare to use the axiom of choice). You would still need to formalize this proof but this is how I convinced myself the claim was true.

Nic
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