A question about powersets

[KyriakosCH]

I have a question, about a property of powersets. Namely:
Can you prove that any element of a set will exist in half of the set's powerset, without using the method in the proof that the number of subsets is always 2^(number of elements in the original set)?
Cause obviously it is trivial to just go "you change one of the 2s to a 1, so you have half".

(yes, I know I am still at AGS, but there's no harm in asking ^_^ )

[Snarky]

Yes, you show (by simple construction) that there's a 1:1 mapping between sets that have the element and sets that don't, so the two powersets with and without the element have the same size.

(A simple way to think of it is as a series of bits: the number of e.g. 8-bit numbers with, say, the fifth bit set is obviously the same as the number of 8-bit numbers with that bit off, because for every number where it's on there's a corresponding one where it is off.)

(This proof works for finite sets; I have no idea whether it generalizes to infinite sets.)

[KyriakosCH]

Yes, you show (by simple construction) that there's a 1:1 mapping between sets that have the element and sets that don't, so the two powersets with and without the element have the same size.

(A simple way to think of it is as a series of bits: the number of e.g. 8-bit numbers with, say, the fifth bit set is obviously the same as the number of 8-bit numbers with that bit off, because for every number where it's on there's a corresponding one where it is off.)

(This proof works for finite sets; I have no idea whether it generalizes to infinite sets.)

Thanks But you already inferred the possible issue with this type of proof. It is based on construction. I am not trying to prove that this is so for any specific set (and for those with few elements it would be easy to just construct the set) but to have another way to generalize on either that half the subsets will always have one element or (equivalent statement) that each element of the original set is there as many times as any other element. These two (particularly so the last...) are intuitively accepted by a human, but I am not interested in intuition.

I am basically trying to establish if any proof (generalization) of the above can exist, without featuring as part of it (or being in tautology with it) the proof that the number of subsets will be 2^n (n number of the elements in the original set). I asked this somewhere else too, so here is my presentation of that proof:

"Given that the number of subsets (including empty and tautology to the original) of any original set with n number of elements is 2^n (this is true because there are only two cases for element x1 to be in the subsets or not, and this gets multiplied by the co-existence or lack of the x2,x3...xn, so it is 2(2)(2)... n times = 2^n), and since if you take one element out, this leads to one of those 2s being transformed to a 1 (cause there is only one way to not be there), it follows the sets which contain (or do not contain) any specific element are 2^n divided by 2."

Can you think of any other proof which doesn't require construction nor does it rely on the above, to prove either of the two sentences?

[Snarky]

You can do a constructive proof "in general," for a set defined only in the abstract. I am confident that the proof I outlined can be formalized to apply to any finite set.

The thing you ultimately want to prove is trivially true by symmetry.

[KyriakosCH]

The thing is, I am not trying to establish if it is true - to prove it. I already provided a proof of it. I am trying to establish if there is any proof which isn't essentially the same as the one I provided - and proof by symmetry is essentially the same.
Anyway, thanks for your reply. At any rate I don't expect there to be an actually different proof. It would be cool if there was, cause then it would be usable along with the other proof, to show other stuff.

[Snarky]

The thing is, I am not trying to establish if it is true - to prove it. I already provided a proof of it. I am trying to establish if there is any proof which isn't essentially the same as the one I provided - and proof by symmetry is essentially the same.

I don't really agree that proof by symmetry is "essentially" the same as a proof that relies on |P(S)| = 2^|S|. But yeah, if you don't want a proof to rely on "an element either is or is not included in a set, and if you count those alternatives you get 2," I think it's going to be tough to do the first part (or to even define what you mean by "half"). Maybe you could generalize it into some more complex domain (e.g. probabilistic sets), prove a more general theorem and then derive the finding as a special case.

For the second part (all elements occur in an equal number of sets in the power set), you can just note that by definition, to construct the power set P(S) from S we first order S "in any order." So to prove that elements e1 and e2 in S occur equally often in the subsets of S (the elements of P(S)), simply swap them in the ordering of S. The sets that before had e1 now have e2, and vice versa. But by definition, P(S) remains the same, so the number of sets with e1 and the number of sets with e2 must both be unchanged, and therefore the two are equal. Since e1 and e2 are arbitrary elements of S, this is true for any two elements, and therefore for all elements.

[KyriakosCH]

I agree with you, we are just not talking about the exact same thing Think of it this way: imagine a computer which didn't intuitively know either the definition or the obvious ties, and was programmed to never use anything which forms the known proof (P(s) and extrapolations).
For a computer it obviously is a usual state to be unaware of ties that to a human are self-evident (nothing new here, and I am not suggesting that you were unaware of that; just making it clearer what I was after )

As for your guess, I agree with it: if there is actually some different proof, it would involve going from a vastly general case (itself a whole different realm) to this.

Edit: One last note. I did like the idea you wrote about just renaming the elements, making use of position not playing a role in powersets. This doesn't have to factor duality, though I am not sure how it can be used along with the duality-based proof (to show other stuff, I mean; tbh I am thinking of uses with formal logic). But maybe it can, since it would also tie this to probability where position does matter, so at least allow for tools specific to that (and search there for something not crucially the same as the P(s)).

Therefore, thank you for your replies!