Axiom of Choice: Disjoint Family ##\Rightarrow ## Power Set

[Terrell]

So apparently the proof involves a trick that converts the problem of a general power set $\mathscr{P}(M)$ of some set $M$ which has of course the property of not having pairwise disjoint set-elements to a problem that involves disjoint set-elements. I do not understand why this trick is valid because I think by doing so, we are then "re-proving" the case where the set-elements are disjoint.

 

[fresh_42]

It would have been helpful to know the precise wording of all three versions, for otherwise we can only guess.

Now, what if the author wouldn't have called it a trick and simply defined the family $\mathcal{F}$? Would you still ask, if this is a valid family in the sense of the theorem?

 

[Terrell]

It would have been helpful to know the precise wording of all three versions, for otherwise we can only guess.

let me put it up in a minute.

 

[Terrell]

Would you still ask, if this is a valid family in the sense of the theorem?

I would think so since it is a defined family $\mathscr{F}$ then it would not be an arbitrary family..?

 

[fresh_42]

I don't see that an arbitrary family is needed. We need an arbitrary power set, which we have. Then we apply the disjoint version on $\mathscr{F}$, for which we do not need arbitrariness. And finally we turn back on what it means for $\mathscr{P}(M)$.

 

[Terrell]

Then we apply the disjoint version on F

So the defined $\mathscr{F}$ is the disjoint version of $\mathscr{P}(M)$?

 

[fresh_42]

As far as I can understand, not knowing the precise definitions of either of them. The structure is as follows:

• To be proven: power set form
• given: any (arbitrary) power set $\mathscr{P}(M)$
• given disjoint family form is true for any disjoint family

• define $\mathscr{F}$
• apply disjoint family form on $\mathscr{F}$: as it is valid for all families of disjoint sets, it is also valid for $\mathscr{F}$
• deduce power set form for $\mathscr{P}(M)$

 

[Terrell]

• define $\mathscr{F}$
• apply disjoint family form on $\mathscr{F}$: as it is valid for all families of disjoint sets, it is also valid for $\mathscr{F}$
• deduce power set form for $\mathscr{P}(M)$

So this "trick" is a set up so that we can formally deduce the power set form. Correct? I guess, I got so fixated in the "jump" in logic as to how the author of the proof have derived the "trick".

 

[fresh_42]

So this "trick" is a set up so that we can formally deduce the power set form. Correct?

Yes. The trick is, that we can only apply the disjoint family form, so we define $\mathscr{F}$ and make it applicable. We then still have to prove the power set form from that.

 

[Terrell]

A bit unrelated. Do you think this trick could have been derived by working backwards? Anyway, thank you!

 

[fresh_42]

I'm not quite sure what you mean by backwards. It's more like "If you are not willing, then I need violence". One can ask: If I only may apply disjoint, but my power set isn't, how can I make it fit? Don't know, whether this can be called backwards, but it is what's going on.