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Is there an axiomatic proof for this? I think it is intuitively correct but I cannot convince myself with proof.

Derek Elkins left SE
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NEo Yang
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    What, precisely, do you mean by an "index set"? I'm particularly interested in knowing why the set itself can't serve as its own index set. – Robert Shore Jul 31 '19 at 19:00
  • I am also confused. From wiki it seems that index set is used to label the element, but labeling elements with themselves do not make much sense. – NEo Yang Jul 31 '19 at 19:06
  • From wikipedia: "an index set is a set whose members label (or index) members of another set". It seems that a set cannot be the index set of itself by definition. – NEo Yang Jul 31 '19 at 19:09
  • If you insist that the index set be different in some way than the set that it indexes... assume without loss of generality that the set we wish to index has $1$ as an element but does not have $0$ as an element. Then the set where we replaced the $1$ with a $0$ is now a distinct set unequal to the original that can act as an index set for it... I don't see why you should insist that the index set be "different" than the original set though. Suppose we wanted the index set for ${0,1,2,3,4,5}$ or for ${1,2,3,4,5,6}$... the most natural and obvious choice would be themselves. – JMoravitz Jul 31 '19 at 19:18
  • I would guess that the ordinal numbers are so uncountably infinite that you can index any set with them, probably with an infinite number of them to spare which is good since they don't actually form a set from my understanding of axiomatic set theory. However, I'm not familiar enough with set theory to attempt a proof – Fady Nakhla Jul 31 '19 at 19:24
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    I agree with @JMoravitz, but if you really want it to be different, index $S$ by $S\times{0}$ – saulspatz Jul 31 '19 at 19:24
  • So the whole point of having them is just for convenient notations so that we can distinguish each of the element? – NEo Yang Jul 31 '19 at 19:25
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    @NEoYang: When you say "another set" is not "a different set". – Asaf Karagila Jul 31 '19 at 19:26
  • @FadyNakhla This assumption is equivalent to the axiom of choice. – Berci Jul 31 '19 at 19:28
  • @saulspatz I was under the impression that the ordinal numbers generalize that natural numbers in an uncountable (but well-ordered) way. Is that not the case? there doesn't seem to be an obvious bijection between the natural and ordinal numbers since the first infinite ordinal is identified with N itself. – Fady Nakhla Jul 31 '19 at 19:30
  • @Berci Do you know of anywhere I could find a proof of their equivalence? – Fady Nakhla Jul 31 '19 at 19:32
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    @FadyNakhla Sorry, somehow, I misread "ordinal numbers" as "natural numbers". Yes, you can index any set with an ordinal (or a cardinal.) – saulspatz Jul 31 '19 at 19:33
  • @Berci Is it due to the fact that indexing with the ordinals is equivalent to the well-ordering theorem, which is itself equivalent to the axiom of choice? Sorry if this is beginning to get off topic, but this might be of interest to the OP if whatever axiomatic system he is using includes that axiom of choice. – Fady Nakhla Jul 31 '19 at 19:43
  • @FadyNakhla yes, it is just the well-ordering theorem. – Berci Jul 31 '19 at 21:55

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