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

  • #1
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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.
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  • #2
fresh_42
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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?
 
  • #3
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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.
 
  • #4
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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..?
 
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  • #5
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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)##.
 
  • #6
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Then we apply the disjoint version on F
So the defined ##\mathscr{F}## is the disjoint version of ##\mathscr{P}(M)##?
 
  • #7
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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)##
 
  • #8
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  • 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".
 
  • #9
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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.
 
  • #10
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A bit unrelated. Do you think this trick could have been derived by working backwards? Anyway, thank you!!
 
  • #11
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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.
 

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