# Is this proof using the axiom of choice correct?

In Munkres' Topology, the axiom of choice is defined as follows:

Given a collection $A$ of disjoint nonempty sets, there exists a set $C$ containing exactly element from each element of $A$; that is a set $C$ such that $C$ is contained in the union of elements of $A$, and for each $a$ in $A$, the intersection of $C$ and $a$ is a singleton set.

Now the text asks to prove the existence of a choice function for an arbitrary collection of non-empty sets. Here's what I did:

For each $a$ in $A$, consider the set $\{a\}$. The axiom of choice as defined above guarantees that we can pick out the so called representative element of the set $a$. Call it $c_a(a)$. We can do this for every $a$ in $A$. So now, we define our choice function $c$ such that for any $a$ in $A$, $$c(a)=c_a(a)$$ This defines the choice function completely and completes the proof.

Is this proof correct, because the proof given in the book is something different and made me wonder if this proof wasn't sophisticated enough. Thanks a lot for your time.

• I don’t understand at all what you mean by “the so called representative element of the set $a$”. The fact that you are considering the set $\{a\}$ (which has one element) suggests thta you are picking $a$ itself which is not a good idea. – k.stm Feb 9 '15 at 16:27
• Also, I want to advertise a different view point on the axiom of choice: I think it should as often as possible be formulated as “Every surjective map of sets $r\colon M → N$ has a right inverse $s\colon N → M$ (that is, $r∘s = \mathrm{id}_N$).” With this formulation, it is also clear what is meant by the choice function. And it avoids talking about “collections of sets” and so forth. – k.stm Feb 9 '15 at 16:33
• @k.stm: No, that is really a nice formulation for category theory, and sure it has merits in set theory as well. From here to "should as often as possible be formulated as" is quite a distance, and I disagree with that. – Asaf Karagila Feb 9 '15 at 17:05
• @AsafKaragila Okay, so this is where I’m coming from: Many people (including myself) get very confused by other formulations (such as Munkres’) which are hard to grasp and hard to apply. The “set-epis are split”-formulation is by far the simplest, most conceptual and handiest I know of. This makes it the best formulation for my day-to-day work as a mathematics student (and I wish I had stumbled upon it sooner). If you are disputing its position as “the best version”, are you refering to set theory? Do you have other candidates or are they all equally good? – k.stm Feb 9 '15 at 17:40
• @k.stm: Yes. Every family of nonempty sets admits a choice function. If you dislike "family", every set of nonempty sets admits a choice function. – Asaf Karagila Feb 9 '15 at 17:43

The formulation of the axiom by Munkres requires the sets to be disjoint, you didn't not define any disjoint sets, or perhaps you have, the $\{a\}$'s, but then you get a set $C$ which chooses from each singleton an element. Not from each set in your family of nonempty sets.