Does all collection of sets have an index-set?

Question

I have this form of the axiom of choice:

Suppose that C is a collection of nonempty sets. Then thre exists a function $f:C\rightarrow \cup_{A\in C}A$ such that $f(A)\in A$ for each $A \in C$.

I have this exercise:

Show that the axiom of choice is equivalent to the following statement : If $\{A_i\}_{i\in I}$ is any indexed collection of nonempty sets, then $\times _{i \in I}A_i \ne \emptyset$.

Now, having the axiom of choice, I think it is ok to show that the second statement follows from the first one. But assuming the second statement, and also assuming that we have the collection C, I have a little trouble showing the axiom of choice. Because I can not just assume that I can index the sets?, and create an index set?, or can I?

Collection of all sets? Takes cover. – Pedro Jul 09 '15 at 03:17 — Pedro, Jul 09 '15 at 03:17
possible duplicate of Can every set be written as a family? – Cameron Buie Jul 09 '15 at 03:34 — Cameron Buie, Jul 09 '15 at 03:34

score 2 · Accepted Answer · answered Jul 09 '15 at 03:18

If you have a set $C$ which you want to index as $\{A_i\}_{i\in I}$, the natural choice would be to set $I=C$ and $A_c=c$. There's no rule that says you can't, and it certainly functions. (And one would only tend to care about the fact that an indexed collection is different from a set if they wished to include some members multiple times)

score 0 · Answer 2 · answered Sep 04 '23 at 23:55

An indexing of a set $C$ over some index set $I$ is a mapping $$ f : I \rightarrow C, \quad i \mapsto c_i $$ There is at least such index set $I$ and mapping $f$: we simply pick $I := C$ and let $f$ be the identity: $$ f : C \rightarrow C, \quad i \mapsto i . $$

Does all collection of sets have an index-set?

2 Answers2