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> But for the set of real numbers, no matter what sequence we choose, we will never ever find the diagonalized number in our sequence. So then it’s only provin

by nico 4y ago

> But for the set of real numbers, no matter what sequence we choose, we will never ever find the diagonalized number in our sequence. So then it’s only proving that if you choose the ordering ahead of time, you won’t be able to do it, because real numbers don’t have a predefined order. You can only order them as you produce them. If you re-sort after each iteration (or insert them in order), you can count as many real numbers as you want. In any case, for practical applications, you could never count anything infinite, at some point you’d run out of physical storage to keep the count. How many bits can be encoded in the universe? That will give the limit of what could ever be counted. And it’s not infinite.

hgsgm 4y ago
We're talking about infinity, not practical considerations. > (or insert them in order), There is no order. You can't describe one, and it is proven that no order exists. > you can count as many real numbers as you want No, you can't count more than 0% of them, even in infinite steps.
- nico 4y ago
  The order is arbitrary. What is the order of natural numbers? If you say they are in ascending order, that’s just the order in which you count them. The same way, I can just generate random real numbers, then define my set as in ascending order, and insert them in that order. Which is what we essentially do with natural numbers as well. Except that the order is a sort of mainstream standard and has been drilled into our minds since babies.
  - LegionMammal978 4y ago
    > The same way, I can just generate random real numbers, then define my set as in ascending order, and insert them in that order. To perform the diagonalization, we can simply look at each real number that you randomly generate and add digits to the diagonalized number based on that order. It doesn't matter how much you shuffle them around afterward: the diagonalized number will still be different from all the generated numbers.
    - nico 4y ago
      What you are saying is: 1) I will produce an infinite amount of numbers, then stop 2) I will look at the infinite amount of numbers and then produce a number which is not within the infinite numbers 3) Hey look, the number I produced in 2) is not within the infinite set, so there must be a larger infinite set The issue is that you can never finish 1). That’s the error/illusion. But if you get to cheat, then I can also add a step 4) As long as you add digits to my numbers to generate an additional number, I will reorder the set of generated real numbers and assign each of them a natural number. Of course the order will change from iteration to iteration. But in this way the generated real numbers will always have a corresponding natural number.
      - LegionMammal978 4y ago
        Where did I imply 1)? We never finish producing the numbers, so unless our production order was fixed from the beginning, we never know the full value of the diagonalized number. But the partial diagonalized number (which we refine at each step by looking at the produced number) will gradually converge to a limit, which is a real number different from any number we will ever produce. At no point in the process do we know the limit exactly, but it is there nonetheless. > Of course the order will change from iteration to iteration. But in this way the generated real numbers will always have a corresponding natural number. The real numbers that you generate will always have a corresponding natural number. But the diagonalized number will never have a corresponding natural number, no matter how much you shuffle your generated values.
        nico 4y ago
        You are saying that when you stop counting, the diagonal number will be different. That means that you stop. Which implies you are not going to infinity. If you are going to infinity, there is no side that “wins” (real or naturals). You can always keep generating an additional one for either side. The same way, we could continue this discussion forever if we just want to keep going back and forth (although HN does impose a limit of replies). And you could only make an assertion about the length of each side of the conversations, after the conversation ends. But if the conversation keeps going forever, you could never say to talked more than the other. Unless you specify a finite range for which you want to measure. Naturals and Reals are both infinite. Saying that the cardinality of one is different than the cardinality of the other, is just comparing the ideas of cardinality about them that we have come with, but their actual cardinality can never be known.
  - haneul 4y ago
    Ascending order is just one of many arbitrary orders. An ordering is a first output of a generator function, a second output, a third output, etc, where for any arbitrarily chosen element X of the set (such as the positive integers or reals), we can show that the generator will reach X in N steps, for some finite N. That is why ascending order is the norm for the positive integers. For any positive integer X, it is clear that we will reach it in X steps using that generator function. So, it’s not a matter of being mainstream, it’s just the simplest. The difference between the infinity of the reals and the positive integers is that there is no such generator function for the reals. No matter what function you give me, I can write a finite number that you will not reach in a finite number of steps. Feel free to give me a generator function and I will give you a finite number that it cannot reach in finite steps. On the other hand, try giving me a finite positive integer that I cannot reach in finite steps.
    - nico 4y ago
      Neither the reals nor the naturals “exist”. We produce them as we need them. Reals and naturals both have the same number of symbols: unknowable (“infinity”). > Feel free to give me a generator function and I will give you a finite number that it cannot reach in finite steps. f(n)=n If you give n to that formula, it will generate n in 1 step. > On the other hand, try giving me a finite positive integer that I cannot reach in finite steps. Can’t come up with one. As long as is finite you can always just output the number itself. Anything finite is countable. Anything infinite is unknowable.
      - haneul 4y ago
        Nono. F(n) where n is the number of steps, not the desired number. You’ve got it backwards. The number is the output, not the input.
        nico 4y ago
        N was the number of steps, but sure: F(x)=x, will give you x in one step.
LegionMammal978 4y ago
> So then it’s only proving that if you choose the ordering ahead of time, you won’t be able to do it, because real numbers don’t have a predefined order. You can only order them as you produce them. What's the problem? You just produce the diagonalized number at the same time as you produce your order. Every order you produce them in has its own corresponding diagonalized number that will never be produced. > If you re-sort after each iteration (or insert them in order), you can count as many real numbers as you want. You can count as many real numbers as you want, but none of them will ever be equal to the diagonalized number. (That is, in finite time, you will always be able to find a digit that differs between the two numbers.) > How many bits can be encoded in the universe? That will give the limit of what could ever be counted. And it’s not infinite. How do you know the universe isn't infinite? Regardless, I don't see why we shouldn't consider formal systems in which infinite amounts of information can be manipulated. Even in our finite light cone, mathematical statements about infinite sets can help us separate the possible from the impossible. For instance, mathematics tells us that adding 1 to any integer will never produce the same integer. There are infinitely many integers, so we can't experimentally verify this in the real world. But if we accept that abstract statement, then we know what to expect when we do try to physically add one object to a group of objects.
haneul 4y ago
If as you say, the infinity of the real numbers lacks the property of predefinable order, whereas the infinity of the natural numbers does have that property, does it not strike you that those infinities may be fundamentally different in some way relating to size, and therefore mapability?
- nico 4y ago
  It only strikes me that we are trying so hard to know the unknowable. Infinite essentially mean unknown. How could you possible compare the size of two unknowns? You just can’t, but we want to be able to reduce something infinite/unknowable, to a finite symbol that we can use in our limited language. Edit: for sibling comment -> how would you know the cardinality of a set without counting it? And if you have two infinite sets, how would you ever finish counting them to determine their cardinality and compare them? It doesn’t even make sense to talk about the cardinality of an infinite set, infinity is not a number. We pretend that the infinity symbol can somehow magically summarize and quantize something that is essentially unknowable. The infinity symbol doesn’t make something magically finite. The cardinality of an infinite set is unknowable. It is not “infinity”.
  - LegionMammal978 4y ago
    > Infinite essentially mean unknown. How could you possible compare the size of two unknowns? The point is, that's not really what infinite means. A set being infinite just means that we can start listing elements, and we'll always be able to find a new element that we haven't listed before. To compare two infinite sets A and B, we can consider functions F that take any element from A and output some element from B. If A and B have the same cardinality, then we can always describe a certain function F that can potentially output any element from B, if we know the right input from A. But if B has a greater cardinality than A, then we cannot find any such function. No matter which function F we choose, there will always be certain values in B that can never be output by F given any input in A. Thus, we say that B has "more elements" than A in a certain sense. In the case of Cantor's argument, A is the set of natural numbers, B is the set of real numbers, and F is the ordering we choose. A variation of Cantor's argument can be expressed fully in terms of finite objects. Suppose we have a function F such that F(i) = f_i, where f_i is a function such that f_i(j) represents the jth digit of the ith real number in the sequence. Then, we can write a new function g(j) = (F(j)(j) + 2) mod 10, or some variation. This function g(j) similarly represents a real number. Now, we can take any i we want and start comparing f_i(1) vs. g(1), f_i(2) vs. g(2), etc. After a finite amount of time, we'll always reach a point where the two functions differ. This means that the two functions represent two different real numbers. Therefore, the real number represented by g is not represented by any of the f_i, no matter which function F we start with.

by nico 4y ago

hgsgm 4y ago

We're talking about infinity, not practical considerations. > (or insert them in order), There is no order. You can't describe one, and it is proven that no order exists. > you can count as many real numbers as you want No, you can't count more than 0% of them, even in infinite steps.

nico 4y ago
The order is arbitrary. What is the order of natural numbers? If you say they are in ascending order, that’s just the order in which you count them. The same way, I can just generate random real numbers, then define my set as in ascending order, and insert them in that order. Which is what we essentially do with natural numbers as well. Except that the order is a sort of mainstream standard and has been drilled into our minds since babies.
- LegionMammal978 4y ago
  > The same way, I can just generate random real numbers, then define my set as in ascending order, and insert them in that order. To perform the diagonalization, we can simply look at each real number that you randomly generate and add digits to the diagonalized number based on that order. It doesn't matter how much you shuffle them around afterward: the diagonalized number will still be different from all the generated numbers.
  - nico 4y ago
    What you are saying is: 1) I will produce an infinite amount of numbers, then stop 2) I will look at the infinite amount of numbers and then produce a number which is not within the infinite numbers 3) Hey look, the number I produced in 2) is not within the infinite set, so there must be a larger infinite set The issue is that you can never finish 1). That’s the error/illusion. But if you get to cheat, then I can also add a step 4) As long as you add digits to my numbers to generate an additional number, I will reorder the set of generated real numbers and assign each of them a natural number. Of course the order will change from iteration to iteration. But in this way the generated real numbers will always have a corresponding natural number.
    - LegionMammal978 4y ago
      Where did I imply 1)? We never finish producing the numbers, so unless our production order was fixed from the beginning, we never know the full value of the diagonalized number. But the partial diagonalized number (which we refine at each step by looking at the produced number) will gradually converge to a limit, which is a real number different from any number we will ever produce. At no point in the process do we know the limit exactly, but it is there nonetheless. > Of course the order will change from iteration to iteration. But in this way the generated real numbers will always have a corresponding natural number. The real numbers that you generate will always have a corresponding natural number. But the diagonalized number will never have a corresponding natural number, no matter how much you shuffle your generated values.
      - nico 4y ago
        You are saying that when you stop counting, the diagonal number will be different. That means that you stop. Which implies you are not going to infinity. If you are going to infinity, there is no side that “wins” (real or naturals). You can always keep generating an additional one for either side. The same way, we could continue this discussion forever if we just want to keep going back and forth (although HN does impose a limit of replies). And you could only make an assertion about the length of each side of the conversations, after the conversation ends. But if the conversation keeps going forever, you could never say to talked more than the other. Unless you specify a finite range for which you want to measure. Naturals and Reals are both infinite. Saying that the cardinality of one is different than the cardinality of the other, is just comparing the ideas of cardinality about them that we have come with, but their actual cardinality can never be known.
- haneul 4y ago
  Ascending order is just one of many arbitrary orders. An ordering is a first output of a generator function, a second output, a third output, etc, where for any arbitrarily chosen element X of the set (such as the positive integers or reals), we can show that the generator will reach X in N steps, for some finite N. That is why ascending order is the norm for the positive integers. For any positive integer X, it is clear that we will reach it in X steps using that generator function. So, it’s not a matter of being mainstream, it’s just the simplest. The difference between the infinity of the reals and the positive integers is that there is no such generator function for the reals. No matter what function you give me, I can write a finite number that you will not reach in a finite number of steps. Feel free to give me a generator function and I will give you a finite number that it cannot reach in finite steps. On the other hand, try giving me a finite positive integer that I cannot reach in finite steps.
  - nico 4y ago
    Neither the reals nor the naturals “exist”. We produce them as we need them. Reals and naturals both have the same number of symbols: unknowable (“infinity”). > Feel free to give me a generator function and I will give you a finite number that it cannot reach in finite steps. f(n)=n If you give n to that formula, it will generate n in 1 step. > On the other hand, try giving me a finite positive integer that I cannot reach in finite steps. Can’t come up with one. As long as is finite you can always just output the number itself. Anything finite is countable. Anything infinite is unknowable.
    - haneul 4y ago
      Nono. F(n) where n is the number of steps, not the desired number. You’ve got it backwards. The number is the output, not the input.
      - nico 4y ago
        N was the number of steps, but sure: F(x)=x, will give you x in one step.

LegionMammal978 4y ago

> So then it’s only proving that if you choose the ordering ahead of time, you won’t be able to do it, because real numbers don’t have a predefined order. You can only order them as you produce them. What's the problem? You just produce the diagonalized number at the same time as you produce your order. Every order you produce them in has its own corresponding diagonalized number that will never be produced. > If you re-sort after each iteration (or insert them in order), you can count as many real numbers as you want. You can count as many real numbers as you want, but none of them will ever be equal to the diagonalized number. (That is, in finite time, you will always be able to find a digit that differs between the two numbers.) > How many bits can be encoded in the universe? That will give the limit of what could ever be counted. And it’s not infinite. How do you know the universe isn't infinite? Regardless, I don't see why we shouldn't consider formal systems in which infinite amounts of information can be manipulated. Even in our finite light cone, mathematical statements about infinite sets can help us separate the possible from the impossible. For instance, mathematics tells us that adding 1 to any integer will never produce the same integer. There are infinitely many integers, so we can't experimentally verify this in the real world. But if we accept that abstract statement, then we know what to expect when we do try to physically add one object to a group of objects.

haneul 4y ago

If as you say, the infinity of the real numbers lacks the property of predefinable order, whereas the infinity of the natural numbers does have that property, does it not strike you that those infinities may be fundamentally different in some way relating to size, and therefore mapability?

nico 4y ago
It only strikes me that we are trying so hard to know the unknowable. Infinite essentially mean unknown. How could you possible compare the size of two unknowns? You just can’t, but we want to be able to reduce something infinite/unknowable, to a finite symbol that we can use in our limited language. Edit: for sibling comment -> how would you know the cardinality of a set without counting it? And if you have two infinite sets, how would you ever finish counting them to determine their cardinality and compare them? It doesn’t even make sense to talk about the cardinality of an infinite set, infinity is not a number. We pretend that the infinity symbol can somehow magically summarize and quantize something that is essentially unknowable. The infinity symbol doesn’t make something magically finite. The cardinality of an infinite set is unknowable. It is not “infinity”.
- LegionMammal978 4y ago
  > Infinite essentially mean unknown. How could you possible compare the size of two unknowns? The point is, that's not really what infinite means. A set being infinite just means that we can start listing elements, and we'll always be able to find a new element that we haven't listed before. To compare two infinite sets A and B, we can consider functions F that take any element from A and output some element from B. If A and B have the same cardinality, then we can always describe a certain function F that can potentially output any element from B, if we know the right input from A. But if B has a greater cardinality than A, then we cannot find any such function. No matter which function F we choose, there will always be certain values in B that can never be output by F given any input in A. Thus, we say that B has "more elements" than A in a certain sense. In the case of Cantor's argument, A is the set of natural numbers, B is the set of real numbers, and F is the ordering we choose. A variation of Cantor's argument can be expressed fully in terms of finite objects. Suppose we have a function F such that F(i) = f_i, where f_i is a function such that f_i(j) represents the jth digit of the ith real number in the sequence. Then, we can write a new function g(j) = (F(j)(j) + 2) mod 10, or some variation. This function g(j) similarly represents a real number. Now, we can take any i we want and start comparing f_i(1) vs. g(1), f_i(2) vs. g(2), etc. After a finite amount of time, we'll always reach a point where the two functions differ. This means that the two functions represent two different real numbers. Therefore, the real number represented by g is not represented by any of the f_i, no matter which function F we start with.