Infinity paradox? Help

[arbiteroftruth]

I suspect I'm missing some simple clause about the way we define things like "infinity", but this apparent paradox occurred to me.

Suppose I want to make a set of all the finite positive integers. I order them as 1, 2, 3, ..., such that each element is necessarily equal to its index within the set.

Since there is no upper bound to finite integers, the size of the set must be infinite. Thus the value of the index must become infinite. Since the value of the index and the value of the corresponding element must be equal, the value of a given element becomes infinite. But we said from the start that this is a set of only finite integers. From this we must conclude that there exist integers that are both finite and infinite.

I understand conceptually that this comes from the difference between an infinite process (adding new elements) and "infinity" as a number, but I can't see how to explicitly define the distinction in such a way that avoids this paradox.

Or do we simply accept the existence of numbers that are both finite and infinite, and recognize that their finiteness becomes irrelevant because their infiniteness makes them impossible to write down as anything other than just "infinity"?

[mfb]

>> Thus the value of the index must become infinite
No. There is an infinite number of finite integers. You just can't give an upper bound for your integers (because there is none).

[arbiteroftruth]

>> Thus the value of the index must become infinite
No. There is an infinite number of finite integers. You just can't give an upper bound for your integers (because there is none).

Perhaps I should phrase it this way:

Given the way I've ordered the set, the number of elements in the set cannot exceed the values that are allowed to exist within the set, because each element of the set is a value corresponding to exactly how many elements are included up to that point. So if the number of elements within the set is infinite, then the set must contain elements with a value of infinity.

[korona]

Consider the function f: N -> Sets: n -> {1, 2, ..., n}.
The function is clearly injective and thus it is a bijection into the image of f. But the set of positive integers is not in the image of f.

Given the way I've ordered the set, the number of elements in the set cannot exceed the values that are allowed to exist within the set, because each element of the set is a value corresponding to exactly how many elements are included up to that point.

This statement is only true for the finite sets of the form {1, 2, ..., n} but not for the whole set of positive integers. There is no reason why this statement should be true for the set of all positive integers.

[zmic]

>> Thus the value of the index must become infinite
No. There is an infinite number of finite integers. You just can't give an upper bound for your integers (because there is none).

Perhaps I should phrase it this way:

Given the way I've ordered the set, the number of elements in the set cannot exceed the values that are allowed to exist within the set, because each element of the set is a value corresponding to exactly how many elements are included up to that point. So if the number of elements within the set is infinite,

"The number of elements in the set is infinite" is somewhat confusing. There is no such thing as an "infinite number". Every number is a perfectly ordinary number, and therefore every number in the set you describe is a perfectly ordinary number, although some of them will be very big indeed.

"The number of elements in the set is infinite" is just shorthand for "When you try to write down all the elements in this set, say 1 per second, you will never finish". It does not mean that the number of elements in the set IS the number "infinite", because there is no such number.

then the set must contain elements with a value of infinity.

[SunAvatar]

Perhaps I should phrase it this way:

Given the way I've ordered the set, the number of elements in the set cannot exceed the values that are allowed to exist within the set, because each element of the set is a value corresponding to exactly how many elements are included up to that point. So if the number of elements within the set is infinite, then the set must contain elements with a value of infinity.

You can phrase it however you want, it's still not true. There is no rule that says an infinitely long ascending list of positive numbers must contain numbers greater than or equal to the length of the list itself. Such a fact happens to be true for finite lists, but that's because finite lists are special.

In fact you have just proven that this property doesn't hold for infinite lists, so until you can prove that it does there's no paradox, just an attempt to use an intuition about 'size' outside its domain of usefulness.

(Slight topic shift: has some version of this property been offered as a definition of 'finite' in some context? An ordinal κ is finite if every set of ordinals with order type κ has successor-of-limit κ or larger?)

[arbiteroftruth]

Ah, I see my problem here. My usual approach to the behavior of infinity in various circumstances is to take a calculus approach, examining the behavior of the scenario as the variable of interest becomes arbitrarily large, and defining the behavior at infinity to be the same as the limit found by increasing the variable. In this particular scenario, what I'm doing is noticing that the arbitrarily large variable always remains finite, and applying this approach to define infinity as finite, even though the whole point was to make the variable become infinite, so I'm simply contradicting my own method.

To add a little extra context though, this is how I got myself into this train of thought:

Any integer can be generated using only the number 1 and a sufficient number of subtraction operations. I considered the notion of defining the set of integers as all numbers that can be generated by this method. I then wondered whether such a definition should specify a finite or infinite process of such subtractions, and I realized that I couldn't find any effective difference. Any integer can be achieved with a finite number of such steps, but because there is no upper bound, the number of steps can be arbitrarily large. To equate that "arbitrarily large" with infinite I now realize is the same misapplication of the calculus approach.

But this all makes me wonder, how usefully can we define infinity beyond those aspects in which it is essentially shorthand for "arbitrarily large"? There is use in comparing sizes of various infinitely large sets, though in that case infinite becomes essentially shorthand for "outside the scope of the number system.

But then, and this is switching topics, that makes me wonder something else about various infinities. The only proof I know of(and there may well be more for all I know) that the real numbers are uncountable while the integers aren't goes something like this. Try to make a list of all the real numbers, and this will be a list of infinitely long strings of digits. I can then take the first digit of the first element in the list, and specifically choose a different digit for that position. I can then take the second digit of the second element, and again specifically choose a different digit for that position. Thus, I can guarantee that there is no element in your list that matches all the digits I have selected, because I have specifically chosen digits to disagree with every element on the list in at least one position. Thus, my digits constitute a valid number that isn't on your list, therefore the initial assumption that you have a list of all the reals has created a contradiction, thus such a list is impossible.

But can't the exact same argument be made for, for example, the integers? I can select a ones digit that disagrees with the first element in your list, a tens digit that disagrees with the second element in the list, and so on, thus guaranteeing that I generate a number that isn't on your list.

And to come at it from the other direction, as I understand it we prove that the set of rational numbers is no greater than the set of integers by finding a method that assigns each rational number to a corresponding integer and guarantees that no rational numbers are missed. Thus we conclude the set of rational numbers cannot be greater than the set of integers.

But can't the same be done for the reals? If every real number can be expressed as an infinite decimal, then I can equate the integers' ones digit to the reals' ones digit, the integers' tens digit to the reals' tenths digit, the integers' hundreds to the reals' tens, the integers' thousands to the reals' hundredths, and so on, and guarantee that all possible digit positions receive all possible values, thus no real number is left out.

What am I missing here?

[ahammel]

But then, and this is switching topics, that makes me wonder something else about various infinities. The only proof I know of(and there may well be more for all I know) that the real numbers are uncountable while the integers aren't goes something like this. Try to make a list of all the real numbers, and this will be a list of infinitely long strings of digits. I can then take the first digit of the first element in the list, and specifically choose a different digit for that position. I can then take the second digit of the second element, and again specifically choose a different digit for that position. Thus, I can guarantee that there is no element in your list that matches all the digits I have selected, because I have specifically chosen digits to disagree with every element on the list in at least one position. Thus, my digits constitute a valid number that isn't on your list, therefore the initial assumption that you have a list of all the reals has created a contradiction, thus such a list is impossible.

But can't the exact same argument be made for, for example, the integers? I can select a ones digit that disagrees with the first element in your list, a tens digit that disagrees with the second element in the list, and so on, thus guaranteeing that I generate a number that isn't on your list.

You can't construct an integer from the diagonal of an infinite matrix because integers have a finite number of digits by definition.

And to come at it from the other direction, as I understand it we prove that the set of rational numbers is no greater than the set of integers by finding a method that assigns each rational number to a corresponding integer and guarantees that no rational numbers are missed. Thus we conclude the set of rational numbers cannot be greater than the set of integers.

But can't the same be done for the reals? If every real number can be expressed as an infinite decimal, then I can equate the integers' ones digit to the reals' ones digit, the integers' tens digit to the reals' tenths digit, the integers' hundreds to the reals' tens, the integers' thousands to the reals' hundredths, and so on, and guarantee that all possible digit positions receive all possible values, thus no real number is left out.

Same problem: you can't have infinitely long integers, so you can't map the integers to the reals like that.

[skullturf]

Try to make a list of all the real numbers, and this will be a list of infinitely long strings of digits.

But can't the exact same argument be made for, for example, the integers? I can select a ones digit that disagrees with the first element in your list, a tens digit that disagrees with the second element in the list, and so on, thus guaranteeing that I generate a number that isn't on your list

No, the exact same argument can't be made for the integers.

This is because, as you note above, each real number is (essentially) an infinitely long string of digits.

But each integer is (essentially) a finite string of digits.

You can't make an integer not in your list by selecting the ones digit a certain way, then selecting the tens digit a certain way, then selecting the hundreds digit a certain way, and so on. The thing that you construct is not guaranteed to be an integer.

(Ninja'd)

[arbiteroftruth]

You can't construct an integer from the diagonal of an infinite matrix because integers have a finite number of digits by definition.
Same problem: you can't have infinitely long integers, so you can't map the integers to the reals like that.

That restriction on integers seems arbitrary though. What contradiction, if any, would arise if I defined a class of objects identical to the integers except that the number of digits is allowed to be infinite? And if there is no contradiction, why do we build that restriction into the definition?

EDIT: Come to think of it, there's no way a contradiction could come out of that, as the set I'm thinking of already exists. It's simply the set of all real numbers in which every digit right of the decimal is 0. So why don't we allow the set of integers to be equal to that set?

[jestingrabbit]

What would fail is mathematical induction ie there would be propositions that are true for 0 and true for n+1 if true for n, but not true for all integers (a trivial example of such a proposition would be "having a finite number of digits in its representation", but other problems, like determining divisibility or primality, would also crop up pretty quickly).

[ConMan]

Try following my general rule of thumb - infinities break things.

There are a huge number of statements S(x) that are true for every finite x but false in the limit as x increases without bound. S(x) = "x is finite" is a nice simple case, but then there are things like S(n) = "f(n) is a rational number" (where f(n) is an integer function that returns, say, the first n digits of pi). Or my favourite one, which always seems to break certain people's brains, S(n) = "0.999...9, where there are n 9s after the decimal place, is strictly less than 1".

[skullturf]

You can't construct an integer from the diagonal of an infinite matrix because integers have a finite number of digits by definition.
Same problem: you can't have infinitely long integers, so you can't map the integers to the reals like that.

That restriction on integers seems arbitrary though. What contradiction, if any, would arise if I defined a class of objects identical to the integers except that the number of digits is allowed to be infinite? And if there is no contradiction, why do we build that restriction into the definition?

EDIT: Come to think of it, there's no way a contradiction could come out of that, as the set I'm thinking of already exists. It's simply the set of all real numbers in which every digit right of the decimal is 0. So why don't we allow the set of integers to be equal to that set?

There's absolutely nothing wrong with considering that set, and wondering what properties it has, or what can be proved about it.

But should we call it the integers?

Roughly speaking, when we talk about the positive integers, we mean 1,2,3,4,5,6,7,8,9,10,... and nothing else. We only call something a positive integer if it can be obtained from 1 by adding 1 finitely many times. This idea is related to jestingrabbit's statement that we want mathematical induction to be true for the positive integers.

By all means, feel free to consider and play around with nonstandard definitions of some of our commonly used sets. But just be aware that sometimes things are considered nonstandard for a reason, and sometimes the observations you're making have been made by others using slightly different language.

[arbiteroftruth]

What would fail is mathematical induction ie there would be propositions that are true for 0 and true for n+1 if true for n, but not true for all integers (a trivial example of such a proposition would be "having a finite number of digits in its representation", but other problems, like determining divisibility or primality, would also crop up pretty quickly).

Would not my original post be an example showing that that's already the case? If the proposition is "the set of integers from 0 to n has exactly n+1 elements in it", this is true when n=0, and true for n+1 whenever it's true for n, but is not true of the set of all integers.

[jestingrabbit]

What would fail is mathematical induction ie there would be propositions that are true for 0 and true for n+1 if true for n, but not true for all integers (a trivial example of such a proposition would be "having a finite number of digits in its representation", but other problems, like determining divisibility or primality, would also crop up pretty quickly).

Would not my original post be an example showing that that's already the case? If the proposition is "the set of integers from 0 to n has exactly n+1 elements in it", this is true when n=0, and true for n+1 whenever it's true for n, but is not true of the set of all integers.

That's not a proposition that's true or false for sets, that's a proposition on the natural numbers, so when you say "but is not true of the set of all integers" I wonder what you could be meaning.

[ahammel]

Come to think of it, there's no way a contradiction could come out of that, as the set I'm thinking of already exists. It's simply the set of all real numbers in which every digit right of the decimal is 0. So why don't we allow the set of integers to be equal to that set?

You can't apply the diagonal argument to that set because as soon as you change one of the zeros trailing the decimal to something else, the result is a number that isn't in the set.

[arbiteroftruth]

You can't apply the diagonal argument to that set because as soon as you change one of the zeros trailing the decimal to something else, the result is a number that isn't in the set.

True, but it doesn't have to be a perfect diagonal. To use the standard example of the reals, the example can be easily modified to not include every digit. I could make my first digit disagree with the first digit of the first element, pick my second digit completely at random, make my third digit disagree with the third digit of the second element, pick my fourth digit at random, and so on. Or I could simply pick the first n digits at random, make my n+1th digit disagree with the n+1th digit of the first element, and so on.

Similarly, if using the set of all reals in which all digits right of the decimal are 0, I simply leave all those positions alone, and start taking a "diagonal" from the ones digit leftward. The result is guaranteed to not match any existing element in the list, and still satisfies the definition of the set in question.

@ jestingrabbit:

Similarly, I'm not sure what you mean either. Could you clarify your objection?

[Nyktos]

Similarly, I'm not sure what you mean either. Could you clarify your objection?

He means, it is true that for all natural numbers n the set {0, 1, ..., n} has n + 1 elements. Letting n equal "infinity" causes problems, but infinity is not a natural number.

[++$_]

Similarly, I'm not sure what you mean either. Could you clarify your objection?

You wrote: "If the proposition is 'the set of integers from 0 to n has exactly n+1 elements in it'...." This is a proposition about the natural number n.

Now, the set N of all natural numbers is not of the form "the set of integers from 0 to n." (Other sets that are not of this form include the set of rational numbers, the set of real numbers, and the set of one-legged barmaids in Switzerland.)

So why would you think your proposition was true about the set of all natural numbers? It doesn't even make sense for that set, because the quantity "n+1" isn't defined when you haven't got an n.

To put it another way, you certainly wouldn't claim it's true of the set of all one-legged barmaids in Switzerland, would you? And if you did, what would it even mean?

[ahammel]

You can't apply the diagonal argument to that set because as soon as you change one of the zeros trailing the decimal to something else, the result is a number that isn't in the set.

True, but it doesn't have to be a perfect diagonal. To use the standard example of the reals, the example can be easily modified to not include every digit. I could make my first digit disagree with the first digit of the first element, pick my second digit completely at random, make my third digit disagree with the third digit of the second element, pick my fourth digit at random, and so on. Or I could simply pick the first n digits at random, make my n+1th digit disagree with the n+1th digit of the first element, and so on.

Similarly, if using the set of all reals in which all digits right of the decimal are 0, I simply leave all those positions alone, and start taking a "diagonal" from the ones digit leftward. The result is guaranteed to not match any existing element in the list, and still satisfies the definition of the set in question.

No it isn't. If you leave element n of the diagonal unchanged, the number constructed from the diagonal isn't guaranteed to be different from the real number at n.

[fishfry]

I suspect I'm missing some simple clause about the way we define things like "infinity", but this apparent paradox occurred to me.

Suppose I want to make a set of all the finite positive integers. I order them as 1, 2, 3, ..., such that each element is necessarily equal to its index within the set.

Since there is no upper bound to finite integers, the size of the set must be infinite. Thus the value of the index must become infinite.

No. The index of 47 is 47. The index of 343434334 is 343434334. The index of 9^9^9^9 is 9^9^9^9.

Can you tell me which element in your sequence fails to have a finite index? In fact every element has a finite index.

[arbiteroftruth]

No it isn't. If you leave element n of the diagonal unchanged, the number constructed from the diagonal isn't guaranteed to be different from the real number at n.

Why should it?

If the digits of a given number are columns, and separate real numbers are rows, what I'm doing is constructing a new row which, when compared against any existing row, differs in at least one column. What you're pointing out merely shows that it won't necessarily differ in all columns, but that was never the goal.

EDIT: To phrase it another way, I could perform the diagonal argument on a supposed list of all real numbers, and then precede every single number on the list, including my new added number, with the digit '1'. My number would have the same first digit as every other number on the list, but the process I used on every other digit beforehand still guarantees that my number is unique. This is no different from simply skipping that first digit to begin with and applying the diagonal starting with the second digit. And to be clear, that's using the second digit of the first element in the list, then the third digit of the second element, and so on. A digit is skipped, but no element is skipped.

[fishfry]

Ah, I see my problem here. My usual approach to the behavior of infinity in various circumstances is to take a calculus approach, examining the behavior of the scenario as the variable of interest becomes arbitrarily large, and defining the behavior at infinity to be the same as the limit found by increasing the variable.

Then your approach to calculus will run into trouble too. Example: Take a sequence whose n-th term is 1/n. So the sequence is 1, 1/2, 1/3, 1/4, ...

Now it's certainly true that each element of the sequence is strictly greater than 0.

Is that true for the limit of the sequence? No, it's not. Properties frequently are NOT preserved by taking limits. So if you're carelessly just assuming that limits preserve all properties you're interested in, you'd get into a lot of trouble.

A more striking example is to start in the lower-left corner of a square and go one unit to the right, one unit up. Total path is 2.

Now instead, start at the lower left and go 1/2 unit right, 1/2 unit up, 1/2 unit right, 1/2 unit up. Now your total path is still 2. If you keep on doing this you'll see that your path appears to converge to the diagonal, which has length sqrt(2). But the limit of the "little to the right, little up..." path is 2.

You can NOT blindly take limits and assume that properties are preserved.

[arbiteroftruth]

Ah, I see my problem here. My usual approach to the behavior of infinity in various circumstances is to take a calculus approach, examining the behavior of the scenario as the variable of interest becomes arbitrarily large, and defining the behavior at infinity to be the same as the limit found by increasing the variable.

Then your approach to calculus will run into trouble too. Example: Take a sequence whose n-th term is 1/n. So the sequence is 1, 1/2, 1/3, 1/4, ...

Now it's certainly true that each element of the sequence is strictly greater than 0.

Is that true for the limit of the sequence? No, it's not. Properties frequently are NOT preserved by taking limits. So if you're carelessly just assuming that limits preserve all properties you're interested in, you'd get into a lot of trouble.

A more striking example is to start in the lower-left corner of a square and go one unit to the right, one unit up. Total path is 2.

Now instead, start at the lower left and go 1/2 unit right, 1/2 unit up, 1/2 unit right, 1/2 unit up. Now your total path is still 2. If you keep on doing this you'll see that your path appears to converge to the diagonal, which has length sqrt(2). But the limit of the "little to the right, little up..." path is 2.

You can NOT blindly take limits and assume that properties are preserved.

The "greater than 0" example is the same basic error I already recognized. The property of being strictly separated from the limit is the very thing I'm trying to avoid by taking the limit in the first place, so to make that error would be to violate the starting assumption of taking a limit at all.

As for the path example, there's no real problem so long as length between two points is defined to be the shortest possible path distance. For me to get the 2 result proves that I can take a path that hits all the same points as the length path but has a longer distance, but it does not redefine the length itself.

[jestingrabbit]

Similarly, I'm not sure what you mean either. Could you clarify your objection?

Sure. I said that having "infinite digit" integers would break mathematical induction. The principle of mathematical induction says that for propositions on the natural numbers, P, if

• P(0), and
• P(n) implies P(n+1)

then P(n) is true for all natural numbers, n.

The proposition that you wrote down was "the set of integers from 0 to n has exactly n+1 elements in it". I took that to be a proposition on the natural numbers, because if its not then mathematical induction doesn't apply to it. From that stance, you can't apply the proposition to the set of all natural numbers, because that isn't an integer. On the other hand, if you want your proposition to be a proposition on some set of subsets of the natural numbers, mathematical induction doesn't directly apply.

So, either you have a proposition that is subject to mathematical induction, and it makes no sense to apply it to the set of naturals, or you have a proposition that isn't subject to mathematical induction, and therefore isn't a counterexample to the principle of mathematical induction.

[arbiteroftruth]

Similarly, I'm not sure what you mean either. Could you clarify your objection?

You wrote: "If the proposition is 'the set of integers from 0 to n has exactly n+1 elements in it'...." This is a proposition about the natural number n.

Now, the set N of all natural numbers is not of the form "the set of integers from 0 to n." (Other sets that are not of this form include the set of rational numbers, the set of real numbers, and the set of one-legged barmaids in Switzerland.)

So why would you think your proposition was true about the set of all natural numbers? It doesn't even make sense for that set, because the quantity "n+1" isn't defined when you haven't got an n.

To put it another way, you certainly wouldn't claim it's true of the set of all one-legged barmaids in Switzerland, would you? And if you did, what would it even mean?

Understood.

I suppose what I'm really arguing is applying induction in a sort of "reverse" way. Rather than performing induction on the elements in question, performing induction on the iteration of whatever process is defined. I hope that makes sense. For example, if I make the proposition "starting with 0 and incrementing by 1 yields an integer equal to the number of incrementations carried out". This holds true if the number of incrementations is 0, and it holds true for n+1 incrementations if it holds true for n incrementations, but it doesn't hold true for infinite incrementations, as then the result is either finite or not an integer.

I realize that's not the same as standard induction, and the way I've constructed it to specifically allow infinite incrementations is what creates the disconnect, but my point is that allowing the integers to be infinite also can be helpful for certain reasoning methods.

[++$_]

The problem is that carrying out a process (such as "increment this") an infinite number of times is not a well-defined concept. Before you can use it, you have to define it. And once you do that, there are no problems any more.

For example, one way to make it defined is to use the ordinal numbers. However, then you get something that is not the natural numbers (instead, it is the ordinals). No paradox arises because to do induction on the ordinal numbers you have to use transfinite induction, and that requires you to check the result for every limit ordinal (not just for a single base case).

The only time anything goes wrong is when you try to apply a theorem about the natural numbers -- the Principle of Mathematical Induction -- when you are dealing with some set that is not the natural numbers, such as the ordinals, or the infinite decimal words.

[arbiteroftruth]

I haven't rigorously proved this of course, but it seems to me it should be possible to create a consistent number system in which behavior "at infinity" is always by definition the behavior found by the limit of making the relevant variable arbitrarily large. Examples such as fishfry brings up can be avoided by careful definitions of objects and their properties (such as a line having a constant slope, for example, avoiding the problem of a staircase approximation), and though that may lead to an ugly amount of revised definitions of things, that strikes me as no more objectionable than having a ton of exceptions to rules specifically for the sake of infinity.

Given such a number system, standard induction on natural numbers would apply to an infinite process as well by the definition of how infinity is treated.

[Talith]

I haven't rigorously proved this of course, but it seems to me...

This is where you should have stopped. Think about these things of course, play about with them, enjoy the head aches, but be careful with statements like the above; they always tend to bite you in the ass.

[arbiteroftruth]

I haven't rigorously proved this of course, but it seems to me...

This is where you should have stopped. Think about these things of course, play about with them, enjoy the head aches, but be careful with statements like the above; they always tend to bite you in the ass.

I don't see how it possibly could, since an admission of uncertainty is built right into the structure of the sentence. But fine, I acknowledge that I have been warned.

[ConMan]

I haven't rigorously proved this of course, but it seems to me...

This is where you should have stopped. Think about these things of course, play about with them, enjoy the head aches, but be careful with statements like the above; they always tend to bite you in the ass.

I don't see how it possibly could, since an admission of uncertainty is built right into the structure of the sentence. But fine, I acknowledge that I have been warned.

It's more the "it seems to me", because vast swathes of things that seem obvious in mathematics are actually, when you get down to it, false (and when they're true, often very hard to prove so). Also, you're proposing a system where you create awkward redefinitions of objects in order to avoid awkward results - which tends to go against lots mathematics where the aim is to start with simple constructs and see what, ugly or elegant, can be proved from them. As a result, you're going to have to ask a very difficult question - what *can't* your system do that existing systems can, because by removing some of the counterintuitive results you're also almost certainly going to remove some more useful (or even important) ones.

[arbiteroftruth]

Also, you're proposing a system where you create awkward redefinitions of objects in order to avoid awkward results - which tends to go against lots mathematics where the aim is to start with simple constructs and see what, ugly or elegant, can be proved from them. As a result, you're going to have to ask a very difficult question - what *can't* your system do that existing systems can't, because by removing some of the counterintuitive results you're also almost certainly going to remove some more useful (or even important) ones.

I could just as easily say the standard system uses an awkward definition of infinity to avoid awkward results about other principles, which goes against the aim of starting with a simple construct for things like infinity and letting the other principles be as ugly or elegant as such a system proves them to be.

Ultimately, it's all just a web of interrelating concepts, and we choose for our own ease of use which concepts will be axiomatic and which will be fleshed out by complex reasoning.

Anyway, with the disclaimer that I obviously don't know for sure, it seems to me that such a system wouldn't cause problems until one gets into very complex branches of mathematics. Convergent series, integration, and differentiation (and thus calculus) are unaffected, because those all depend on the type of treatment of infinity I'm advocating anyway. Induction would extend to infinite numbers by definition, and even if that creates difficulties, one can still use standard induction to draw conclusions about numbers of any given arbitrary size as required by the application. Even many set theory applications that depend on one infinite set always being greater than another infinite set could, I suspect, be retained without much difficulty by defining a construction algorithm that guarantees the one set is larger than the other for all finite iterations, at which point this system would say the same holds true by definition when the set sizes become infinite. Things like, for example, mapping all integers to all even integers, can easily be re-phrased as one algorithm that increments by 1 and another algorithm that increments by 2, each adding one new element with each iteration thus each always being equal in size.

For examples like the stairstep approximation of a line, though my system would have to more carefully define a line, this is no more complex than the standard system's approach of carefully defining when you can or can't treat a limit approaching infinity as an actual result "at" infinity. In the particular example of the stairstep, it's basically only a semantic difference.

The only real issue that comes to mind so far is the whole thing that started this train of thought, namely the fact that I can simultaneously construct a set guaranteed to contain every possible real number and construct an algorithm for finding a real number not contained on that set. The standard system's distinctions between countable and uncountable infinities avoid this issue. In my system, I think I would have to resort to citing the Incompleteness Theorem to write off the latter algorithm as long-hand for "something not included in this system". I at least find that quirk of my system no more objectionable or counterintuitive than the standard system's quirk that integers simultaneously have no upper bound yet are by definition finite.

EDIT: It might also be possible to argue that the diagonal construction of a number not contained in the set of real numbers is just a special case of the notion that a given set of elements can be constructed at different rates, similar to constructing even numbers at the same rate as natural numbers, and related to the concept that, if one were to think of infinity as a number, infinity = infinity +1. The diagonal construction is constructing a new number in which the rate at which new digit positions are updated is linear while the rate at which new positions are updated in the construction of all real numbers is logarithmic. That is, the new number being constructed has, at each iteration, a digit position being modified that wasn't modified at any previous step, while the construction of the set of all real numbers modifies one digit ten times before modifying the next digit once. Hence, in terms of updated new digits, the algorithm constructing a new number is perpetually way ahead of the algorithm for constructing the set of all real numbers. This would imply the need for a formal definition of a concept like "data compression" within my system, to capture the notion that a construction of all real numbers can implicitly contain every real number while being fundamentally unable to explicitly list almost any of them.

I really am curious now about what the pros and cons of such a system would be.

EDIT 2: Though it occurs to me now that such a concept of "data compression", particularly since I bring it up specifically with regard to what can meaningfully be listed, would almost certainly be functionally identical to the standard system's distinctions between arbitrarily large, countably infinite, and uncountably infinite. On the other hand, if that's the case, then this system would almost certainly be just as usable as the standard system, with essentially just a difference in language, which would be the aspect in which all the pros and cons lie. I'll have to think about this more.

[MartianInvader]

Hey, I really like the idea of integers with an infinite number of digits! Apparently, so do some other people.

Forget the .99999... = 1 arguments. I want to argue about ...999999 = -1!

[gmalivuk]

at least find that quirk of my system no more objectionable or counterintuitive than the standard system's quirk that integers simultaneously have no upper bound yet are by definition finite.

The set of integers has no upper bound, but each element in that set is finite. Is this distinction still not clear to you?

[arbiteroftruth]

at least find that quirk of my system no more objectionable or counterintuitive than the standard system's quirk that integers simultaneously have no upper bound yet are by definition finite.

The set of integers has no upper bound, but each element in that set is finite. Is this distinction still not clear to you?

It's perfectly clear and has been for some time. It's just at least as odd as the result I described within my hypothetical system.

[lightvector]

I'm interested to hear how your hypothetical system might dodge the following paradox:

Consider the property "is not infinite". Clearly, 0 is not infinite, 1 is not infinite, 2 is not infinite, ..., so taking the limit, the property "is not infinite" should also be true of infinity. But that's contradictory and silly. And it's not restricted to just numbers. Various versions of this argument can apply to sets or whatever other objects are in your system that come in both finite and infinite versions, if you insist on properties always holding true in the limit.

There are, of course, several ways to get around this issue. But the question is how much you have to give up in the process. I'm curious: arbiteroftruth, what do you think are the best ways to prevent this paradox from occurring in your system?

[fishfry]

at least find that quirk of my system no more objectionable or counterintuitive than the standard system's quirk that integers simultaneously have no upper bound yet are by definition finite.

The set of integers has no upper bound, but each element in that set is finite. Is this distinction still not clear to you?

It's perfectly clear and has been for some time. It's just at least as odd as the result I described within my hypothetical system.

Why? In the set {0,1,2,3,4} there are 5 elements, but each element of the set is less than 5. Why should a set have the same properties as its members? In the set {0,1,2,3,...} there are infinitely many members, each of which is finite. In my grocery bag there are 6 oranges, but my paper bag is not an orange. And none of the oranges are made of paper. What is the problem with distinguishing between the properties of a set and the properties of its individual elements?

[arbiteroftruth]

@lightvector

That's a matter of properly defining the operation of taking the limit. That is, by definition, taking the limit as the variable approaches infinity implies explicitly replacing the property "is finite" with the property "is infinite", and my system would propose that, in doing so, all other properties are determined by how they converge, if at all.

@fishfry

"Wrong" with it? Nothing. As for counterintuitive and quirky, it is such because of the very situation I outlined in my first post. For something like the integers, it's easy to construct a set in which, for all finite sizes, that particular set does have similar properties to its members, but this principle does not hold when the construction is carried out infinitely.

[gmalivuk]

That's a matter of properly defining the operation of taking the limit. That is, by definition, taking the limit as the variable approaches infinity implies explicitly replacing the property "is finite" with the property "is infinite"

Fine, but that only gets you out of that one particular paradox. What about the property, "is even (x)or odd"? 1 is even xor odd, 2 is even xor odd, ... n is even xor odd, but their limit is neither. In fact, I can think of very few typical properties of integers that could possibly be applicable to infinity. "Is 0 or 1 or strictly less than its square", "is 0 or has a unique prime factorization (modulo sign)", "is congruent to K modulo N" for any integer choice of K and any N with |N|>1, and on and on and on.

[SunAvatar]

In addition to the above objections, even if you create some self-consistent system of arbiteroftruth-integers, some of which have infinitely many digits, I can still say "consider the set of all arbiteroftruth-integers with finitely many digits" and once again have an infinite set, all of whose elements are finite. Unless you want to posit that some finite numbers are also infinite (what would this even mean?), or alternately declare that certain sets are inadmissible. The latter is actually viable--it's more or less how the hyperintegers work--but it's up to you to give a precise rule for when a set is or isn't admissible. So this extension doesn't actually solve the problem (not that there is any problem to be solved in the first place).