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You have to use a very funny notion of ⊂ for that to make sense. Mathematicians recognize that there is a lot of funny business going on behind the scenes… if y

You have to use a very funny notion of ⊂ for that to make sense. Mathematicians recognize that there is a lot of funny business going on behind the scenes… if you use the typical construction of these sets, then ℤ ⊂ ℚ is for sure false under the set-theoretic definition of what subsets are. So you are going to get a lot of funny explanations of why you can still say “ℤ ⊂ ℚ” and what that means. And what is “the same as” anyway?

hansvm 2y ago
The typical construction is just to show that they _can_ be constructed. Once you have "primitive" integers, rationals, ..., you pick (for this finite example) the largest of them, and define the "integers" to be a subset with a certain property. That's also well-defined, and you find _exactly_ the same integers in the rationals and reals.
interroboink 2y ago
Could you provide an example? I'm not seeing it. Is there some integer that is not also a rational number?
- klodolph 2y ago
  How about this example: the natural number 0 is not a member of the integers, from a set theoretic perspective. In the natural numbers, 0 = {} However, {} is not an element of the integers. This is not something I expect to be easy to understand. This is the standard set theoretic definition of integers that mainstream mathematicians use. This is not some esoteric, fringe theory.
  - Affric 2y ago
    Isn’t the empty set a subset of all sets? A quick look at Wikipedia indicates that there exist other constructions.
    - klodolph 2y ago
      Yes, the empty set is a subset of all sets. I think some wires got crossed somewhere because whether zero is a subset of some other set doesn’t fit into the questions we are trying to answer. You could pick a construction where the natural numbers are a subset of the integers. This is trivial, but this is a poor strategy overall, because you can always find a bigger set of numbers to work with. You can’t take the “biggest” set of numbers and then define all other sets of numbers as subsets of that. It would be kind of like trying to count down from infinity.
      - Affric 2y ago
        I see where I have gone wrong. Element vs subset. Classic error when thinking fast. Thank you for the thought provoking comments.
      - SAI_Peregrinus 2y ago
        The Surreal Numbers manage to be a sort of "biggest" ordered field, in NBG, though they're a proper class not a set. All other ordered fields are subfields of the Surreals. Of course "ordered" is doing a lot, since it excludes the complex numbers, vectors, bivectors, etc. Whether elements of some object that isn't an ordered field should be considered "nubmers" is related but different question.
    - enugu 2y ago
      Yes, {} is a subset(⊂) of all sets, but it is not a member(∊) of all sets. For instance, {} is not a member of {{{}}}. In the Von-Neumann definition, 1:={0}={{}}. So, 0∊1, but 0 is not a member of {1} which is the above set. https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers https://en.wikipedia.org/wiki/Set-theoretic_definition_of_na...
  - interroboink 2y ago
    Ah, thank you. It does actually make some sense, having looked into it some more now (: It looks like the approach of defining ℤ in terms of ℕ is much more tedious to deal with overall, so can see the advantages.
- Cu3PO42 2y ago
  Conceptually, yes, every integer is also a rational number. But they are represented as different objects. It's a bit like saying every int32 is also a double. Yes, every value of int32 fits into a double, but the bit pattern is different. The canonical construction for rational numbers is pairs (a, b) which we interpret as a/b. An integer k "is the same as" (k, 1). So it might be more correct to say every integer has the same value as some rational number. Of course this distinction is pointless most of the time, so we don't worry about it. This is not unique to integers and rationals. It also applies to naturals and integers, rationals and reals, etc.
  - pyrolistical 2y ago
    But an int32 1 isn’t the same as a mathematic object 1. So of course your analogy doesn’t work
    - klodolph 2y ago
      By definition, analogies are between things that aren’t the “same thing”.
    - Cu3PO42 2y ago
      And a double 1.0 isn't the same as a rational 1/1. But I also didn't claim either of these things.
  - _nalply 2y ago
    I understand that this might be a problem in a programming language, even if you decide to ignore that mathematical sets have infinitively many elements. When I programmed an RPN calculator supporting quotients and complex numbers, I had to grapple with the fact that 1 (one) is a quotient (with denominator 1) and a complex number (with imaginary part 0). I solved it by defining the functions and operators taking a certain type and coercing the arguments to that type. One example, the logarithm can take complex numbers. If I had a quotient argument, I coerce it to a complex number before passing it to the logarithm. However I assumed that this is only a problem in programming languages. I am a bit surprised that mathematics also seems to be affected. I am going to study it a bit.
- deleted 2y ago
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deleted 2y ago
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sebtron 2y ago
I don't think it is a "funny" notion of inclusion, but rather that once ℚ is defined, ℤ is silently redefined to be that one subset of ℚ that was used in the construction.
- klodolph 2y ago
  There’s a very deep problem with that—every time you invent a “superset”, do you then have to redefine the subset to be a subset of that “superset”? There is an infinite chain of supersets of the rational numbers, real numbers, etc. Think of it like this… we can define 0, and then define 1 as the successor. Repeating this, we can have a definition for every finite number. But we cannot do this the other way around. We cannot start with ∞, define the predecessor to ∞, and then somehow get back to 0. In other words, if you want to work backwards and say that smaller sets (like the natural numbers) are a subset of the bigger sets (like complex numbers), then you have to pick a “biggest set” containing all numbers, which is unsatisfactory. Somebody always wants a bigger set.
  - ysofunny 2y ago
    > There’s a very deep problem with that—every time you invent a “superset”, do you then have to redefine the subset to be a subset of that “superset”? I have thought about this too, and I'd initially agree with you. but I thought at some point how mathematical history is not extremely dissimilar from this. put in very rough terms: at first humans discovered/invented numbers (i.e. the counting numbers); these started at number one — the first number. later on, at some point we had to go back and realize that there was a zero before number one which "silently" redefined the first number as zero and this created the natural numbers a the modern set-based N edit: adding this alternative rendering of my intended comment triggered by a condescending reply: "mathematics silently redefines stuff all the time. deal with it"
    - klodolph 2y ago
      There is a different answer here which is more satisfactory… which is to use notions of equality other than set theoretic equality. Which is what the article is talking about.
ysofunny 2y ago
I disagree but I'm being nitpicky I think the real funky business happens right between ℚ ⊂ ℝ but maybe all I'm really trying to say is "hey, look at me, I understand how all of ℕ ⊂ ℤ ⊂ ℚ" I'm very very sure that it's possible to give a fully ZF set-theoretic construction of the rationals using sets. in fact IMO the annoying thing is that there are in fact two trivially equivalent constructions that I can think of the top of my head, and I find that even worse than there being none at all
- klodolph 2y ago
  What do you disagree with? Traditionally, you define ℚ as equivalence classes of pairs of integers. The integer 0 is an integer, it’s not an “equivalence class of pairs of integers” and therefore it’s not a rational, in the set-theoretic sense. Real numbers have lots more constructions. https://en.m.wikipedia.org/wiki/Construction_of_the_real_numbers https://en.m.wikipedia.org/wiki/Construction_of_the_real_num...
  - ysofunny 2y ago
    so yes, to answer one of my own questions. I was only really saying that I don't really know any of those constructions; but know all the "preceding" ones. I gotta figure out dedekind cuts or at least learn tarski's fixed point theorem which I'm sure that would allow me to much better understand tarski's construction all in the slow lifelong process of understanding and learning to draw post's lattice finally, to answer your question, I disagree with saying that there's funky business between ℤ ⊂ ℚ because of my alleged claim that there is at least one construction which avoids the problem you describe, but as I was trying to say, these would have a non-unique way to construct number zero which nobody likes
  - zeroonetwothree 2y ago
    You could construct the larger set and then define the smaller sets as specific subsets. For example Conway’s surreal numbers is a fun construction. If you do it that way then there is only one version of “1”
  - anon291 2y ago
auggierose 2y ago
No funny notion of ⊂ at all, it is about how you construct ℚ, for example. You construct Q(ℤ), show that ℤ is embedded in Q(ℤ) via a monomorphism I, and then define ℚ as (Q(ℤ) \ I(ℤ)) ∪ ℤ. For example. You are right that nobody cares about how this works in detail. It does assume though some notion of ℚ and ℤ as collections, but not necessarily sets from ZFC.
anon291 2y ago
Not at all. This is only a problem when doing formal verification due to how the theorem provers work (based on lambda calc). N is the set of zero and all its successors. Z is N extended with all additive inverses, which means it includes all of N. Q is Z with all multiplicative inverses, so it includes all Z. R is the set of Q plus all supremums and infinums of all bounded sets of Q. R is only not Q when you define R with dedekind cuts, which is a nice formal construction, but doesn't change what R is, up to isomorphism.
- nyssos 2y ago
  > up to isomorphism Which is exactly the sort of more sophisticated notion of equivalence you end up needing in these cases.
  - auggierose 2y ago
    See my other comment. You can define R(ℚ) as Dedekind cuts, see that there is a monomorphism I from ℚ to R(ℚ), and then set ℝ = (R(ℚ) \ I(ℚ)) ∪ ℚ. Then you have ℚ ⊂ ℝ and ℝ ≅ R(ℚ). No problem. Except of course if you work in a theorem prover that doesn't support a construction like that.
  - anon291 2y ago
    Yeah, I'm very familiar. The 'up to isomorphism' applies to dedekind cuts et al. I'm pointing out that the nature of R is independent of its definition, and that any mathematical formal system you invent to construct R (up to isomorphism) is just a description. R exists independently on its own. The constructions of R are useful, because you can show things about them that could be applicable to R. However, the article gives a rhetorical question of whether you should be able to say 0 is in 3. No you should not, because while there are many constructions of the natural numbers, the universal property they all share (that there is a zero and a successor) does not permit any notion of inclusion. The construction of the naturals as sets would permit such notions, but these are not relevant to any discussion of the natural numbers. Put in homotopy type theory terms, the natural numbers are the equivalence class of all representations of the natural numbers, and all propositions over the natural numbers are the equivalence classes of all propositions over all representations of the natural numbers. Since one cannot draw an equivalence between '0 \in 3' and any proposition from another construction of the natural numbers (I believe), then the term '0 \in 3' is not a proposition over the natural numbers. Put another way, I posit that there are constructions isomorphic to the natural numbers whose propositions cannot be mapped onto the space of all propositions over the construction of the natural numbers as sets. Whereas, all constructions of the natural numbers will permit a proof of the associativity of addition, or the infinity of primes. So these proofs are what should be accepted as 'proofs' over the natural numbers, but '0 \in 3' would not, because it's missing in many systems. '0 \in 3' could be used in the set representation to be the definition of less than, but it is not a statement that applies to all representations. Is that clearer?
deleted 2y ago
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diffeomorphism 2y ago
You would usually use something like \hookrightarrow instead. The set is isomorphic to a subset. In the same way the real line is "included" in the plane (\R \times \{0\}) or the sphere is not only a subset of the whole space but a submanifold. Math uses lots of polymorphism and overloaded notation, but that is also part of the beauty, where "A=A" can actually be a deep result about concepts being compatible.
paulddraper 2y ago
> You have to use a very funny notion of ⊂ for that to make sense ?? Are you claiming that ℤ ⊄ ℚ This thread is wild.
- gus_massa 2y ago
  There is a subset of Q that is canonicaly isomorph to Z. Probably the distiction is too technical unless you want to do some weird advanced algebra. (Note that in most weird advanced math the distiction doen't matter.) So in most case, mathematicians and non mathematicians just write Z ⊂ Q and live happily ever after. But if you get supertechnical you should use scare quotes Z "⊂" Q or to look more proffesional Z ↪ Q (because tha inclusion is an injective funcion, and sometimes it's useful to think aboout it as a function.). --- To add some confussion: Imagine that you buy in the supermarket a copy of Z that is green and another copy of Z that is red. 1(red) + 2 (red) = 3(red) 1(green) + 2(green) = 3(green) (If I get super technical, I have to define a +(red) operation and a +(green) operation. And probably also a =(red) and =(green) as the article discuss.) Are they the same Z or just canonicaly isomorph or it doen't matter? --- To add even more confussion: You have a copy of abstract Z and a copy inside the rational written as n/1: 1 + 2 = 3 1/1 + 2/1 = 3/1 Are they the same Z or just canonicaly isomorph or it doen't matter?
  - paulddraper 2y ago
    > There is a subset of Q that is canonicaly isomorph to Z. Actually, the statement ℤ ⊂ ℚ reads "The set of integers is a proper subset of rationals."

hansvm 2y ago

The typical construction is just to show that they _can_ be constructed. Once you have "primitive" integers, rationals, ..., you pick (for this finite example) the largest of them, and define the "integers" to be a subset with a certain property. That's also well-defined, and you find _exactly_ the same integers in the rationals and reals.

interroboink 2y ago

Could you provide an example? I'm not seeing it. Is there some integer that is not also a rational number?

klodolph 2y ago
How about this example: the natural number 0 is not a member of the integers, from a set theoretic perspective. In the natural numbers, 0 = {} However, {} is not an element of the integers. This is not something I expect to be easy to understand. This is the standard set theoretic definition of integers that mainstream mathematicians use. This is not some esoteric, fringe theory.
- Affric 2y ago
  Isn’t the empty set a subset of all sets? A quick look at Wikipedia indicates that there exist other constructions.
  - klodolph 2y ago
    Yes, the empty set is a subset of all sets. I think some wires got crossed somewhere because whether zero is a subset of some other set doesn’t fit into the questions we are trying to answer. You could pick a construction where the natural numbers are a subset of the integers. This is trivial, but this is a poor strategy overall, because you can always find a bigger set of numbers to work with. You can’t take the “biggest” set of numbers and then define all other sets of numbers as subsets of that. It would be kind of like trying to count down from infinity.
    - Affric 2y ago
      I see where I have gone wrong. Element vs subset. Classic error when thinking fast. Thank you for the thought provoking comments.
    - SAI_Peregrinus 2y ago
      The Surreal Numbers manage to be a sort of "biggest" ordered field, in NBG, though they're a proper class not a set. All other ordered fields are subfields of the Surreals. Of course "ordered" is doing a lot, since it excludes the complex numbers, vectors, bivectors, etc. Whether elements of some object that isn't an ordered field should be considered "nubmers" is related but different question.
  - enugu 2y ago
    Yes, {} is a subset(⊂) of all sets, but it is not a member(∊) of all sets. For instance, {} is not a member of {{{}}}. In the Von-Neumann definition, 1:={0}={{}}. So, 0∊1, but 0 is not a member of {1} which is the above set. https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers https://en.wikipedia.org/wiki/Set-theoretic_definition_of_na...
- interroboink 2y ago
  Ah, thank you. It does actually make some sense, having looked into it some more now (: It looks like the approach of defining ℤ in terms of ℕ is much more tedious to deal with overall, so can see the advantages.
Cu3PO42 2y ago
Conceptually, yes, every integer is also a rational number. But they are represented as different objects. It's a bit like saying every int32 is also a double. Yes, every value of int32 fits into a double, but the bit pattern is different. The canonical construction for rational numbers is pairs (a, b) which we interpret as a/b. An integer k "is the same as" (k, 1). So it might be more correct to say every integer has the same value as some rational number. Of course this distinction is pointless most of the time, so we don't worry about it. This is not unique to integers and rationals. It also applies to naturals and integers, rationals and reals, etc.
- pyrolistical 2y ago
  But an int32 1 isn’t the same as a mathematic object 1. So of course your analogy doesn’t work
  - klodolph 2y ago
    By definition, analogies are between things that aren’t the “same thing”.
  - Cu3PO42 2y ago
    And a double 1.0 isn't the same as a rational 1/1. But I also didn't claim either of these things.
- _nalply 2y ago
  I understand that this might be a problem in a programming language, even if you decide to ignore that mathematical sets have infinitively many elements. When I programmed an RPN calculator supporting quotients and complex numbers, I had to grapple with the fact that 1 (one) is a quotient (with denominator 1) and a complex number (with imaginary part 0). I solved it by defining the functions and operators taking a certain type and coercing the arguments to that type. One example, the logarithm can take complex numbers. If I had a quotient argument, I coerce it to a complex number before passing it to the logarithm. However I assumed that this is only a problem in programming languages. I am a bit surprised that mathematics also seems to be affected. I am going to study it a bit.
deleted 2y ago
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deleted 2y ago

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sebtron 2y ago

I don't think it is a "funny" notion of inclusion, but rather that once ℚ is defined, ℤ is silently redefined to be that one subset of ℚ that was used in the construction.

klodolph 2y ago
There’s a very deep problem with that—every time you invent a “superset”, do you then have to redefine the subset to be a subset of that “superset”? There is an infinite chain of supersets of the rational numbers, real numbers, etc. Think of it like this… we can define 0, and then define 1 as the successor. Repeating this, we can have a definition for every finite number. But we cannot do this the other way around. We cannot start with ∞, define the predecessor to ∞, and then somehow get back to 0. In other words, if you want to work backwards and say that smaller sets (like the natural numbers) are a subset of the bigger sets (like complex numbers), then you have to pick a “biggest set” containing all numbers, which is unsatisfactory. Somebody always wants a bigger set.
- ysofunny 2y ago
  > There’s a very deep problem with that—every time you invent a “superset”, do you then have to redefine the subset to be a subset of that “superset”? I have thought about this too, and I'd initially agree with you. but I thought at some point how mathematical history is not extremely dissimilar from this. put in very rough terms: at first humans discovered/invented numbers (i.e. the counting numbers); these started at number one — the first number. later on, at some point we had to go back and realize that there was a zero before number one which "silently" redefined the first number as zero and this created the natural numbers a the modern set-based N edit: adding this alternative rendering of my intended comment triggered by a condescending reply: "mathematics silently redefines stuff all the time. deal with it"
  - klodolph 2y ago
    There is a different answer here which is more satisfactory… which is to use notions of equality other than set theoretic equality. Which is what the article is talking about.

ysofunny 2y ago

I disagree but I'm being nitpicky I think the real funky business happens right between ℚ ⊂ ℝ but maybe all I'm really trying to say is "hey, look at me, I understand how all of ℕ ⊂ ℤ ⊂ ℚ" I'm very very sure that it's possible to give a fully ZF set-theoretic construction of the rationals using sets. in fact IMO the annoying thing is that there are in fact two trivially equivalent constructions that I can think of the top of my head, and I find that even worse than there being none at all

klodolph 2y ago
What do you disagree with? Traditionally, you define ℚ as equivalence classes of pairs of integers. The integer 0 is an integer, it’s not an “equivalence class of pairs of integers” and therefore it’s not a rational, in the set-theoretic sense. Real numbers have lots more constructions. https://en.m.wikipedia.org/wiki/Construction_of_the_real_numbers https://en.m.wikipedia.org/wiki/Construction_of_the_real_num...
- ysofunny 2y ago
  so yes, to answer one of my own questions. I was only really saying that I don't really know any of those constructions; but know all the "preceding" ones. I gotta figure out dedekind cuts or at least learn tarski's fixed point theorem which I'm sure that would allow me to much better understand tarski's construction all in the slow lifelong process of understanding and learning to draw post's lattice finally, to answer your question, I disagree with saying that there's funky business between ℤ ⊂ ℚ because of my alleged claim that there is at least one construction which avoids the problem you describe, but as I was trying to say, these would have a non-unique way to construct number zero which nobody likes
- zeroonetwothree 2y ago
  You could construct the larger set and then define the smaller sets as specific subsets. For example Conway’s surreal numbers is a fun construction. If you do it that way then there is only one version of “1”
- anon291 2y ago

auggierose 2y ago

No funny notion of ⊂ at all, it is about how you construct ℚ, for example. You construct Q(ℤ), show that ℤ is embedded in Q(ℤ) via a monomorphism I, and then define ℚ as (Q(ℤ) \ I(ℤ)) ∪ ℤ. For example. You are right that nobody cares about how this works in detail. It does assume though some notion of ℚ and ℤ as collections, but not necessarily sets from ZFC.

anon291 2y ago

Not at all. This is only a problem when doing formal verification due to how the theorem provers work (based on lambda calc). N is the set of zero and all its successors. Z is N extended with all additive inverses, which means it includes all of N. Q is Z with all multiplicative inverses, so it includes all Z. R is the set of Q plus all supremums and infinums of all bounded sets of Q. R is only not Q when you define R with dedekind cuts, which is a nice formal construction, but doesn't change what R is, up to isomorphism.

nyssos 2y ago
> up to isomorphism Which is exactly the sort of more sophisticated notion of equivalence you end up needing in these cases.
- auggierose 2y ago
  See my other comment. You can define R(ℚ) as Dedekind cuts, see that there is a monomorphism I from ℚ to R(ℚ), and then set ℝ = (R(ℚ) \ I(ℚ)) ∪ ℚ. Then you have ℚ ⊂ ℝ and ℝ ≅ R(ℚ). No problem. Except of course if you work in a theorem prover that doesn't support a construction like that.
- anon291 2y ago
  Yeah, I'm very familiar. The 'up to isomorphism' applies to dedekind cuts et al. I'm pointing out that the nature of R is independent of its definition, and that any mathematical formal system you invent to construct R (up to isomorphism) is just a description. R exists independently on its own. The constructions of R are useful, because you can show things about them that could be applicable to R. However, the article gives a rhetorical question of whether you should be able to say 0 is in 3. No you should not, because while there are many constructions of the natural numbers, the universal property they all share (that there is a zero and a successor) does not permit any notion of inclusion. The construction of the naturals as sets would permit such notions, but these are not relevant to any discussion of the natural numbers. Put in homotopy type theory terms, the natural numbers are the equivalence class of all representations of the natural numbers, and all propositions over the natural numbers are the equivalence classes of all propositions over all representations of the natural numbers. Since one cannot draw an equivalence between '0 \in 3' and any proposition from another construction of the natural numbers (I believe), then the term '0 \in 3' is not a proposition over the natural numbers. Put another way, I posit that there are constructions isomorphic to the natural numbers whose propositions cannot be mapped onto the space of all propositions over the construction of the natural numbers as sets. Whereas, all constructions of the natural numbers will permit a proof of the associativity of addition, or the infinity of primes. So these proofs are what should be accepted as 'proofs' over the natural numbers, but '0 \in 3' would not, because it's missing in many systems. '0 \in 3' could be used in the set representation to be the definition of less than, but it is not a statement that applies to all representations. Is that clearer?

deleted 2y ago

[deleted]

diffeomorphism 2y ago

You would usually use something like \hookrightarrow instead. The set is isomorphic to a subset. In the same way the real line is "included" in the plane (\R \times \{0\}) or the sphere is not only a subset of the whole space but a submanifold. Math uses lots of polymorphism and overloaded notation, but that is also part of the beauty, where "A=A" can actually be a deep result about concepts being compatible.

paulddraper 2y ago

> You have to use a very funny notion of ⊂ for that to make sense ?? Are you claiming that ℤ ⊄ ℚ This thread is wild.

gus_massa 2y ago
There is a subset of Q that is canonicaly isomorph to Z. Probably the distiction is too technical unless you want to do some weird advanced algebra. (Note that in most weird advanced math the distiction doen't matter.) So in most case, mathematicians and non mathematicians just write Z ⊂ Q and live happily ever after. But if you get supertechnical you should use scare quotes Z "⊂" Q or to look more proffesional Z ↪ Q (because tha inclusion is an injective funcion, and sometimes it's useful to think aboout it as a function.). --- To add some confussion: Imagine that you buy in the supermarket a copy of Z that is green and another copy of Z that is red. 1(red) + 2 (red) = 3(red) 1(green) + 2(green) = 3(green) (If I get super technical, I have to define a +(red) operation and a +(green) operation. And probably also a =(red) and =(green) as the article discuss.) Are they the same Z or just canonicaly isomorph or it doen't matter? --- To add even more confussion: You have a copy of abstract Z and a copy inside the rational written as n/1: 1 + 2 = 3 1/1 + 2/1 = 3/1 Are they the same Z or just canonicaly isomorph or it doen't matter?
- paulddraper 2y ago
  > There is a subset of Q that is canonicaly isomorph to Z. Actually, the statement ℤ ⊂ ℚ reads "The set of integers is a proper subset of rationals."