# Finding a unique structure of the domain of a function that gives a unique intuitive average?

• Alessandro Iraci
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This question (or similar ones) seems to be recurring. Can I ask you what's motivating you? I find it a bit odd that you are spending so much money for something I struggle to perceive as well defined. Maybe if I understand what you are looking for exactly I can help, but in the current state I simply don't understand the question.

• Alessandro Iraci
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It's actually worse; if I understand correctly, there is no "natural" way to get what you are looking for. Countable sets can be arbitrarily ugly (heck, there are countable models of ZFC!), finding a uniform way to enumerate them it's hardly possible. It can be done with some extra hypotheses (e.g. having a uniform choice function on the universe), but I don't think it will give you anything natural in the sense you're asking for.

• Bharathk98
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AlessandroIraci, perhaps it’s not possible. But there should be cases of A where there is a uniform way to enumerate it. I think my choice function does the job.

• Bharathk98
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I will list examples where we can enumerate as uniformally as possible.

• Alessandro Iraci
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I mean, any example in which you have a set with a "natural" enumeration works, that's pretty much the condition you're asking for, if I understand correctly. The problem is, this cannot be done in general without something like the Axiom of global choice, and even then the result will not be constructive.

• Bharathk98
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What about the choice function listed in my paper. I believe, if possible, there exists an enumeration that gives an infimum to my choice function. I will also list another example (besides the rationals) tomorrow or the day after.

• Alessandro Iraci
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That works only in a few specific examples, and even then, it's an arbitrary choice. As you noticed, different choices of the "approximating" sets F_t give you different results. If you pick a countable set with no "natural" ordering (e.g. the set of all finite permutations), then you can't get a "natural" mean, except in some special cases. I can prove such statements, but I still don't understand what a satisfactory answer would be for you.

• Bharathk98
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For me, a satisfactory answer is finding a “natural mean” for the greatest class of countably infinite sets. If my choice function isn’t enough what choice function could do better. I was hoping to apply the choice function to some examples once I’m older and understand enough advanced maths.

• Alessandro Iraci
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What I am trying to tell you is that there is no such thing, and provably so. There are some weaker, non-constructive alternatives if you assume the Axiom of global choice, or if you loosen the question (e.g. a "natural" mean for sets that come equipped with an enumeration).

• Alessandro Iraci
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Example, consider A=N (the natural numbers) and f to be the identity. How would you define an average here? What if f(n) = (-1)^n * n? Depending on the choice of the F_t sets you can get any real number as an answer.

• Alessandro Iraci
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And this is not even addressing the main problem of having sets with no sensible enumeration: take A to be the set of finite subsets of the set of finite subsets of N (this is constructive - there's stuff that's a lot worse), and f to be any injective function such that the image is { (-1)^n * n | n \in N }. You can literally get any answer depending on the choice of F_t, and there is no choice that is intrinsically better than the others.

• Bharathk98
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Yes but if A=N, using my choice function, we get F_t is any natural number between the interval [-t,t] as the interval grows larger. There are no other options otherwise my choice function would give a larger value for the infimum.

• Bharathk98
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Nevermind you are right. I don’t expect there to always be an average but I know there are cases of function f and A where an average can exist. We will loosen the question to A where an enumeration exists.

• Alessandro Iraci
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I think it can just be restated as asking whether the terms of a series can be arranged in such a way that it converges. If you ask your function to have positive values, then it gets easier. But even with extra conditions (e.g. bounded values) you don't have a unique answer.

• Alessandro Iraci
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For example, let A=N and f(n) = 0 if n is even, 1 if n is odd. Depending on how you choose the F_t, I think you can get pretty much any number between 0 and 1 (for sure you can get any rational, but I think you can get irrationals as well).

• Alessandro Iraci
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I guess the "natural mean" in that case would be 1/2, and if you say f(n) = 0 if n is divisible by some given integer d, and 1 otherwise, you would expect 1/d. But all these functions are essentially the same up to a permutation of the indices.

• Alessandro Iraci
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What you are asking for is to simultaneously enumerate all countable sets. This is roughly equivalent to the axiom of global choice I mentioned before. I think your best shot is to assume V=L (the constructability axiom) and use the induced well-ordering on the universe to get what you are hoping for. But again, this is far from intuitive. If that is enough of an answer for you, I can make this formal.

• Bharathk98
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Perhaps you can show why my choice function of F_t only works for specific cases as well as make my choice function and average rigorous. I think my answer is the most intuitive for a limited number of cases.

• Bharathk98
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I will try and find these “specific cases” myself. I think some example are sets and functions where my choice function gives a unique average. (Note all my questions are in section 0.3)

• Alessandro Iraci
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My point is that I don't think your argument can be made rigorous. The concept of "intuitive" is really not mathematical. To be entirely honest, I don't find your suggested solution very intuitive, and even if it were, its scope is limited to a very small number of cases. What I can do is to show a bunch of counterexamples where different "intuitive" options give different results, and then giving a general rule (not constructive - it won't make you happy) using the axiom V=L.

• Alessandro Iraci
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What I want to stress is that my issue is not that I don't know how to define an average in the sense you are looking for, but the fact that it cannot be done in full generality. If you want a general result, it won't be constructive. If you want a constructive result, I think you have to ask something along the lines A=N, F_t = {1, 2, ... t}, and then there will be some conditions on the function if you want it to have an average. I am sorry, but anything better than this simply can't be done.

• Alessandro Iraci
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Notice also that some of your requests don't apply to generic countable sets. For example, in section 0.2 and in the definition of d(F_n, A), you are implicitly using the fact that A is a subset of the reals, or at the very least that it has a "subtraction" function AxA -> R. The generic countable set does not have this property, nor anything similar (e.g. the set of finite permutations). This invalidates almost every question in 0.3 except the fifth (to which I sort of answered in the comments)

• Alessandro Iraci
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I think your approach works for a countable, bounded subset of the reals (if it is unbounded you have other problems - you can't "spread the numbers evenly"), but it should seldom give a meaningful answer: I expect that for f generic enough you won't get convergence to a mean. And it's likely that some values will be skipped at all, which is probably undesired.

• Alessandro Iraci
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Anyway, this is my offer: I can give you some counterexamples to show you that what you're asking is impossible (most of them I already mentioned in the comments), and a sketch of how to get a uniform approach assuming V=L. If that is fine, I'll take the question, otherwise you can make a counteroffer, but in either case I strongly suspect that the answer will not be what you are looking for. If that is not ok, I will leave it to someone else, but I don't think you'll get anything satisfactory.

• Bharathk98
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Show how my approach works for a countable bounded subset of the reals and make my average and choice function rigorous for this case. If you can do that I will accept you answer.

• Bharathk98
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If f is generic enough it’s true the average will never converge but there must be cases where the average does infact converge. That’s enough to me for a meaningful answer.

• Alessandro Iraci
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You still have the problem that it won't, in general, converge (even in the example you provided with Q \cap [0,1], I'm quite sure one can define f in a way such that your averages have no limit). Moreover, I'm having second thoughts on the "countable bounded subsets of the reals" part, they can be messy. What would you expect to happen if you have an isolated point very far away? It makes all the distances spike up in an unfixable way.

• Alessandro Iraci
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Another question: what would you expect if A = {-1} \cup {1/n | n in N}, and f(x) = 1/x? And without the -1?

• Bharathk98
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Alessandro, I understand but there are example of for Q\cap[0,1] where the average does converge. Consider f(x)=1 when x=(2n+1)/(2m) and f(x)=0 when x=(m)/(2n+1). For my choice of F_t the average does infact converge. As t grows larger the average should be 1.

• Alessandro Iraci
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Yes, there are indeed examples where it converges (say, f = 1 constant...), but there are others in which it does not (f(p/q) = 0 if the minimum n such that q|n! is even, 1 if is odd - this has no limit, it oscillates between 0 and 1 faster and faster), and I don't think there's a meaningful way to classify the functions that do work. It's really an arbitrary condition, it does not generalise.

• Alessandro Iraci
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I thought about it a bit more. I think that your idea only works for intervals of rational numbers (as it is - no extra work to extend it). But still, it will only converge for a restricted family of functions, and I do not think that the condition can be rewritten as anything easier than "it converges". Maybe (maybe!) if f extends to a continuous (or differentiable) function on the corresponding closed interval for the reals. But I expect you will get the integral mean in that case.

• Alessandro Iraci
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Actually, the integral mean can give you a meaningful answer for any countable, dense subset of an interval of the reals, and for any f that is restriction of a continuous function (probably just integrable function) on the corresponding closed interval.

• Alessandro Iraci
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No, sorry, integrable is not sufficient. You need continuity, otherwise every function on a countable set extends to an integrable function (the values it assumes on a countable set are irrelevant). The condition you want is that f extends to a continuous function on the closed interval. Then I suspect any "intuitive average" will give you the same answer, namely the integral mean.

• Alessandro Iraci
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Let me know if this is enough of an answer.

• Bharathk98
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Sure, this will work but I think f need not be continuous. My peice-wise example wasn’t continuous anywhere but I believe an average exists. Perhaps you can clarify on this.

• Alessandro Iraci
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Well, continuity is a sufficient condition, it is not necessary. But again, your example only works for a specific family of F_t, if you assume continuity I think it should work for pretty much every family as long as they are "spaced well".

• Alessandro Iraci
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In your example, there should be a choice of the family F_t that gives an average of 0, and probably other choices that give any real between 0 and 1. Continuity should be enough to fix this.

• Alessandro Iraci
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Waiting for your thumbs up, specifying just in case.

• Bharathk98
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Answers can be viewed only if
1. The questioner was satisfied and accepted the answer, or
2. The answer was disputed, but the judge evaluated it as 100% correct.
• Alessandro Iraci
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Just a small favour: if you are happy with the answer, please do not wait the whole week to accept it, as booking it drained my available credit and I cannot book any other question without money. No hurry, take your time to review the answer, of course, just accept it when you are sure you are satisfied with it. Thank you!

• Bharathk98
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When you say f extend to a continuous function, what is the continuous extension of the function defined on the rationals where f(x)=1 when x is defined with a odd numerator and even denominator and f(x)=0 when x is defined with a even and odd numerator and odd denominator?

• Bharathk98
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I believe I found a case that does not have a continuous extension but used mathematics and found the average can exist. Suppose we have function f defined on values between 0 and 1 and have an even/ odd numerator and odd denominator such that f(x)=0 when x is defined on even numerator and odd denominator and f(x)=1 on an odd numerator and odd denominator. Using wolfram mathematics I get that the answer is 1/2.

• Alessandro Iraci
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Such f does NOT admit a continuous extension, I already mentioned it in the comments. I understand that some of these functions might have an average, but such average will depend on the choice of the family F_t. We agreed to only deal with the case when f does admit a continuous extension.

• Alessandro Iraci
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I don't know what you mean by "using Wolfram", using it how? With that function you can get any number between 0 and 1, depending on the choice of F_t.

• Bharathk98
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It doesn’t matter, since you decided to answer my question when f has a continuous extension this is more than enough.

• Bharathk98
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By “using wolfram” I found a code using Mathematica that allowed me to find a specific average using my choice function. My choice function should give a specific group of F_t which the same average for the examples I listed.

• Bharathk98
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For example for the first example my choice function for F_t gives {m/(n!):m

• Alessandro Iraci
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Oh ok, I see. My point is that, for a different choice of F_t, you get different results, and I don't think there is a natural one in general. I circumvented the problem by requiring continuity, which forces "reasonable" choices to give the same result.

• Alessandro Iraci
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Thank you for accepting the answer quickly! If there is anything else you want to discuss in the comments I remain available.