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Not all infinities can be indexed. Take all the real numbers between 0 and 1, for example. If you pick one and call it N, there's an infinite quantity of real

Not all infinities can be indexed. Take all the real numbers between 0 and 1, for example. If you pick one and call it N, there's an infinite quantity of real numbers between 0 and N and between N and 1. Therefore it's impossible to assign an index to N. Now take rational numbers, which are a subset of real numbers. There's an infinite number of those between 0 and 1 as well, but because it's a subset, there are "fewer" of them. Edit: It appears my sloppy language has ruffled some mathematical feathers. My apologies.

firstlink 3y ago
> If you pick one and call it N, there's an infinite quantity of real numbers between 0 and N and between N and 1. Therefore it's impossible to assign an index to N. This isn't "sloppy language", it's just wrong. Don't try to downplay spreading misinformation. It remains true that there are infinitely many rationals in (0, N) and (N, 1), and yet it is entirely possible to enumerate the rationals (i.e. assign indices). > There's an infinite number of those between 0 and 1 as well, but because it's a subset, there are "fewer" of them. Also misinformation. (0, 0.5) is a subset of (0, 1) yet the two sets have the same cardinality, whether we take them both in Q or both in R.
- roarcher 3y ago
  > This isn't "sloppy language", it's just wrong. Yes, no less than four people before you have explained what's wrong with my comment. I made the "sloppy language" edit before I fully understood them. > Don't try to downplay spreading misinformation. > Also misinformation. LMAO. Every incorrect thing you see on the internet isn't the work of a Russian troll or whatever you're trying to accuse me of here. Sometimes people are just mistaken. Your comment actually has some interesting points that could add to the discussion, but it's ruined by the impotent nerd rage.
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roywiggins 3y ago
That doesn't explain why rationals and reals are different. Rationals are a countable subset of the reals, to prove that you need to produce a method of enumerating them, not just say "they're a subset so they're smaller." Lots of subsets of the reals are uncountable, most notably if the rationals are countable and you split the reals into rationals and irrationals, the irrationals have to be uncountable. Otherwise you'd be forced to say that the union of two countable sets could be uncountable, which isn't true (if you have two countable sets you can automatically produce an enumeration of the union).
- roarcher 3y ago
  I don't see how your point about countability invalidates what I said. Irrationals are uncountable, but there are fewer of them between 0 and 1 than reals between 0 and 1, by virtue of the fact that all irrationals are reals but not all reals are irrationals. I admit I'm using "subset" in the colloquial sense, meaning "a set that's missing some of the elements in its superset", which is not quite the correct definition (a subset doesn't have to be missing anything, i.e., every set is a subset of itself). So technically a subset does not imply fewer elements. Is that what you're getting at? I'm not particularly well versed in set theory, so maybe I'm just misunderstanding your point about countability.
  - roywiggins 3y ago
    You didn't say anything entirely incorrect, I think! But the important bit when it comes to talking about sizes of infinity is that, while both irrationals and rationals are smaller than the reals in one sense (they're both proper subsets), they're very different in another sense: you actually can index the rationals by the natural numbers whereas you can't do the same with the irrationals. That's the much more interesting sense in which the rationals are smaller (or, alternatively, that reals are bigger). The rationals are smaller than the irrationals in this sense as well, despite not being a subset of them at all.
    - roarcher 3y ago
      I think I still don't understand 100%, but that's probably my lack of formal knowledge about this stuff. I appreciate the explanation.
contravariant 3y ago
Depends what you mean by 'indexed'. If you accept the axiom of choice then you can index the real numbers with some uncountable ordinal.
- roarcher 3y ago
  I meant it in the sense that I assume GP meant it, i.e., the index of an element in an ordered set is an integer corresponding to its position in that set, starting at 0 or 1 for the first element (depending on your indexing scheme of choice).
- Dylan16807 3y ago
  Does the axiom of choice let you repeatedly pick the same real numbers in the same order out of a set? Or do you need uncountably infinite storage to keep your list of indexes?
  - contravariant 3y ago
    That's an interesting question, but kind of hard to answer. Programming is inherently constructive so approaching a non-constructive proof as a programming problem tends to be problematic. I don't think you run into any problems if you let the axiom of choice pick the same numbers every time. It would be a bit problematic if you were to add that as an extra axiom (suddenly every proof let x,y in X then ... therefore x=y would become trivial).

firstlink 3y ago

> If you pick one and call it N, there's an infinite quantity of real numbers between 0 and N and between N and 1. Therefore it's impossible to assign an index to N. This isn't "sloppy language", it's just wrong. Don't try to downplay spreading misinformation. It remains true that there are infinitely many rationals in (0, N) and (N, 1), and yet it is entirely possible to enumerate the rationals (i.e. assign indices). > There's an infinite number of those between 0 and 1 as well, but because it's a subset, there are "fewer" of them. Also misinformation. (0, 0.5) is a subset of (0, 1) yet the two sets have the same cardinality, whether we take them both in Q or both in R.

roarcher 3y ago
> This isn't "sloppy language", it's just wrong. Yes, no less than four people before you have explained what's wrong with my comment. I made the "sloppy language" edit before I fully understood them. > Don't try to downplay spreading misinformation. > Also misinformation. LMAO. Every incorrect thing you see on the internet isn't the work of a Russian troll or whatever you're trying to accuse me of here. Sometimes people are just mistaken. Your comment actually has some interesting points that could add to the discussion, but it's ruined by the impotent nerd rage.
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roywiggins 3y ago

That doesn't explain why rationals and reals are different. Rationals are a countable subset of the reals, to prove that you need to produce a method of enumerating them, not just say "they're a subset so they're smaller." Lots of subsets of the reals are uncountable, most notably if the rationals are countable and you split the reals into rationals and irrationals, the irrationals have to be uncountable. Otherwise you'd be forced to say that the union of two countable sets could be uncountable, which isn't true (if you have two countable sets you can automatically produce an enumeration of the union).

roarcher 3y ago
I don't see how your point about countability invalidates what I said. Irrationals are uncountable, but there are fewer of them between 0 and 1 than reals between 0 and 1, by virtue of the fact that all irrationals are reals but not all reals are irrationals. I admit I'm using "subset" in the colloquial sense, meaning "a set that's missing some of the elements in its superset", which is not quite the correct definition (a subset doesn't have to be missing anything, i.e., every set is a subset of itself). So technically a subset does not imply fewer elements. Is that what you're getting at? I'm not particularly well versed in set theory, so maybe I'm just misunderstanding your point about countability.
- roywiggins 3y ago
  You didn't say anything entirely incorrect, I think! But the important bit when it comes to talking about sizes of infinity is that, while both irrationals and rationals are smaller than the reals in one sense (they're both proper subsets), they're very different in another sense: you actually can index the rationals by the natural numbers whereas you can't do the same with the irrationals. That's the much more interesting sense in which the rationals are smaller (or, alternatively, that reals are bigger). The rationals are smaller than the irrationals in this sense as well, despite not being a subset of them at all.
  - roarcher 3y ago
    I think I still don't understand 100%, but that's probably my lack of formal knowledge about this stuff. I appreciate the explanation.

contravariant 3y ago

Depends what you mean by 'indexed'. If you accept the axiom of choice then you can index the real numbers with some uncountable ordinal.

roarcher 3y ago
I meant it in the sense that I assume GP meant it, i.e., the index of an element in an ordered set is an integer corresponding to its position in that set, starting at 0 or 1 for the first element (depending on your indexing scheme of choice).
Dylan16807 3y ago
Does the axiom of choice let you repeatedly pick the same real numbers in the same order out of a set? Or do you need uncountably infinite storage to keep your list of indexes?
- contravariant 3y ago
  That's an interesting question, but kind of hard to answer. Programming is inherently constructive so approaching a non-constructive proof as a programming problem tends to be problematic. I don't think you run into any problems if you let the axiom of choice pick the same numbers every time. It would be a bit problematic if you were to add that as an extra axiom (suddenly every proof let x,y in X then ... therefore x=y would become trivial).