Axiom of choice and function with empty codomain

Question

I'm having a little problem here, namely if the axiom of choice (Wikipedia) is

$$\forall X \left[ \emptyset \notin X \implies \exists f: X \to \bigcup X \quad \forall A \in X \, ( f(A) \in A ) \right]$$

and I choose the nonempty $X=\{\emptyset\}\neq \emptyset$ for which only $\emptyset\in X$ and hence $A=\emptyset$, then I get

$$\exists f:\{\emptyset\}\to\emptyset\ \ \text{such that}\ \ f(\emptyset) \in \emptyset.$$

The last point seems wrong in itself though. Of course, there is no actual value $f(\emptyset)$, so $f(\emptyset) \in \emptyset$ doesn't say much. It might be just taken to be a legal statement. I haven't seen that written down anywhere though.

I don't think $\emptyset\notin{\emptyset}$ is true (see the left hand side of $\Rightarrow$). An implication where the left hand side is false is automatically true. — celtschk, May 28 '13 at 08:42
@celtschk: You're right, thanks. I somehow read the condition to be $\emptyset \neq X$. — Nikolaj-K, May 28 '13 at 08:47

score 1 · Accepted Answer · answered May 28 '13 at 08:48

1

Note that $X=\{\varnothing\}$ satisfies the statement vacuously, so it's fine. You can change the statement to the following:

$$\forall X\exists f\left[f\colon X\to\bigcup X\cup\{\varnothing\}\land\forall A\in X(f(A)\in A\lor A=\varnothing)\right]$$

It allows $\varnothing\in X$, and therefore $X=\{\varnothing\}$. But it's harder to understand it this way, so it's easier to just require $\varnothing\notin X$.

answered May 28 '13 at 08:48

Asaf Karagila

393,674

k, thanks for the addition information. Is there a reason you prefer \varnothing over \emptyset and is there really a point to \colon? – Nikolaj-K May 28 '13 at 08:55
Nick, $\varnothing$ over $\emptyset$ is just a typographical choice. \colon over:is a choice between a separator and a symbol, note that $f:X$ (:) and $f\colon X$ (\colon`) have different spacings. – Asaf Karagila May 28 '13 at 08:56
I know, I saw someone correcting Terry Tao on his blog once, and he thanked that guy. I don't see why $f\colon X$ would be better than $f:X$ though, I feel the second one is somewhat clearer. – Nikolaj-K May 28 '13 at 09:20
TeX.SE: Using \colon or : in formulas? – Martin Sleziak May 28 '13 at 10:21

Axiom of choice and function with empty codomain

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