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This question is best asked with an example. Consider the real numbers. However we construct the real numbers, the "final product" so to speak, is not just a set, but it is a complete ordered field. In other words, it's not just $\mathbb{R}$ but $(\mathbb{R}, +, \cdot, \leq)$. This gives me a sense that the structure associated with $\mathbb{R}$ is intrinsic to it. Yet it isn't; we're free to change the structure however we want.

Now, the way I reconcile this (and I may even be answering my own question!) is that a set is defined by the membership relationship $\in$ and only that. Thus, after constructing $(\mathbb{R}, +, \cdot, \leq)$ (via dedekind, cauchy, whatever), if $x$ is a member of this ordered field that we've created, we define a corresponding set $\mathbb{R}$ such that set theoretically $\mathbb{R} = (\mathbb{R}, +, \cdot, \leq)$ but we don't equip this set with any structure.

Even if this is right, it just feels weird to me and I would like a better or simpler explanation. How exactly does a set with an underlying structure "precede" the set without any structure? It seems "unnatural" in a sense. It's as if the very existence of the reals as a set is contingent on their (standard) ordering and field properties.

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  • $\begingroup$ Of course the structure does not "precede" the set. It is just that we construct $\mathbb R$ to have a handy example. By chance, as a consequence of the means used in the construction, it also comes enabled with some unnecessary structure. What else could we do? Go like "let's imagine a set greater than $\aleph_0$..."? $\endgroup$ – Ivan Neretin Jun 27 '16 at 8:37
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    $\begingroup$ In my opinion, this is a deep question that AFAIK no one yet has an answer to. Personally, I ponder this at least once a month. $\endgroup$ – goblin Jun 27 '16 at 8:54
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    $\begingroup$ @goblin Can you elaborate how you interpret OP's question? I seem to fail to see its depth. $\endgroup$ – Stefan Mesken Jun 27 '16 at 10:18
  • $\begingroup$ @Stefan, sure. I've expanded my comment into a Brutally speculative answer, with a capital 'B'. $\endgroup$ – goblin Jun 27 '16 at 10:44
  • $\begingroup$ It depends a bit on your construction. When you construct $\mathbb{R}$ as the Cauchy completion of $\mathbb{Q}$, you get the elements, namely the set of equivalence classes of Cauchy sequences of rationals under the equivalence relation "$x_n \sim y_n$ iff $x_n-y_n \to 0$". Although $\mathbb{R}$ does wind up inheriting the operations from $\mathbb{Q}$, this is actually part of the definition; we could work just with this "bare" set of equivalence classes if we really wanted to. $\endgroup$ – Ian Jun 27 '16 at 11:05
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The following is mostly just my personal opinion, please keep this in mind.

It is really pointless to think everything in mathematics is a set. It is true (in classical set theory), but this view does not help you at all, unless you are studying set theory specifically.

By that logic it is also somewhat pointless to think of the set of real numbers $\{1,1/2, \pi, \sqrt{2}/2, \dots\}$, because this set is really like any other old set of the same cardinality. These symbols, like $1$, $1/2$, $\pi$ do not have (or should not have!) any meaning in the absence of the ordered field structure of the reals. You are completely right: the structure is not intrinsic to any of these sets of cardinality $|\mathbb{R}|$.

In the same sense, as an example: the game "Sudoku" has nothing to do with the numbers $1,\dots, 9$ at all, because in this game you never add them or compare their size. Instead, you only use any old $9$-element set. It does not matter which one, as they are really all the same. More precisely, they are isomorphic. I am under the impression, that many people realize that isomorphism is the right kind of "equality" for groups and rings and so on, but somehow do not take this seriously enough in the case of sets.

The right kind of logic for this philosophy is type theory. In type-theoretic foundations we have a type of reals as opposed to just a set of reals, which comes with all the structure you want to have. Set theory is like type theory with a single type: that of a set.

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  • $\begingroup$ $\{\varnothing\}$ and $\{\{\varnothing\}\}$ are isomorphic but in standard set theory are unequal. Essentially, do you believe that the definition of "equality" in set theory should be changed? $\endgroup$ – MathematicsStudent1122 Jun 27 '16 at 10:16
  • $\begingroup$ Well... no. I think classical set theory should be completely abolished as foundation of mathematics ducks away. - but yes. These sets are pretty much the same to me (all the structure you can define on of these you can define on the other one too). In the category of sets you say, that they are both terminal objects and terminal objects are unique up to unique isomorphism. That's definitely "equal enough" to me. Again, this is just my philosophy, you don't need to agree with it. $\endgroup$ – Stefan Perko Jun 27 '16 at 11:01
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I'm not sure if I've understood the point you're getting at, but if we have a set $X$ and a bijection $f:X \to \mathbb R$, $f$ will induce a structure $(X, +, \cdot, \leq)$ on $X$, just by using $f$ to identify $x$ with $f(x)$.

Different bijections will give rise to different structures on $X$. So the structure certainly isn't intrinsic to $X$ (or $\mathbb R$), but once we've fixed a "canonical" copy of $\mathbb R$ (with all its extra structure) and a bijection $X \to \mathbb R$ we have a distinguished structure on $X$ induced by this bijection.

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Maybe I'm wrong, but I think this is rather a deep question. Or at least, I ponder it frequently, perhaps once a month. I don't have an answer, but let me just speculate out loud.

Lets just pretend, for a moment, that we kind of accept finitism, just a little bit. We accept finitism a little bit.

Now $\mathbb{N}$ is the initial pointed monounary algebra. What this means is that:

  • $0$ is a natural number,
  • If we've constructed a natural number $n$, then we can construct its successor $S(n)$.

From a finitistic viewpoint, that's as far as we can go; you can only write down finitely many natural numbers at any one time. In some sense, we might say: the thing that makes $\mathbb{N}$ infinite is the function $S$ together with the fact that this constructor is injective. So if we forget $S$, then $\mathbb{N}$ should collapse to a finite set, almost like we've just popped a balloon, and the whole thing just collapses under the pressure of being an initial object. So, we're not allowed to forget $S$. The $S$ is what makes $\mathbb{N}$ infinite.

But here's the thing. Suppose I'm interested in the function $$f : \mathbb{N} \rightarrow \mathbb{N}$$ $$n \mapsto 2\cdot n$$

Well, this isn't a homomorphism of pointed monounary algebras, since $$2 \cdot S(n) = 2\cdot (n+1) = 2\cdot n+2 \neq 2\cdot n+1 = S(2\cdot n)$$

So to define $f$, we have to "forget" that $\mathbb{N}$ is a pointed monounary algebra, and just treat it as a set. In other words, we should really be writing: $$f : U\mathbb{N} \rightarrow U \mathbb{N},$$ where $U$ is the underlying set functor.

But hang on - the underlying set functor "forgets" $S$. Yet somehow, $\mathbb{N}$ fails to collapse under its own weight; it doesn't suddenly shrink to oblivion or anything like that.

Maybe its just me, but I find this a little mysterious. It kind of weirds me out.

Now lets forget finitism. We're back to the good ol' usual way of thinking about math.

Nonetheless, there's still something that kind of weirds me out here, perhaps even more so than the non-existence of a balloon popping effect. Its this: the way we're meant to characterize things is up to unique isomorphism; but, once we've done that, we're allowed to throw away the structure that allowed us to do this, even if this means that the resulting is object is only characterized up to not-necessarily unique isomorphism.

For example, given two sets $X$ and $Y$, we can define $X \times Y$ as the terminal cone to the diagram $$\{0,1\} \rightarrow \mathbf{Set}$$ $$\{0 \mapsto X, 1 \mapsto Y\}$$

This pins down $X \times Y$ up to unique isomorphism.

The problem with this is that we've just defined $X \times Y$ as a cone to a diagram; its not a set. Of course, it has an underlying set, also denoted $X \times Y$, but now suddenly weird we've only got $X \times Y$ characterized up to not-necessarily unique isomorphism.

Are we, like, even allowed to do this?

What does it even mean to be "allowed" to do something in math?

See what I mean? This is, like, deep stuff. Pretty soon you're asking questions like "Why is it that the Mathematic even Exists?" and pondering God and stuff. Its pretty intense. Nonetheless, I suspect that at least some of these musings lead to genuine math (as opposed to e.g. mere philosophy.)

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  • $\begingroup$ Last week I gave a short talk about set theoretical foundations of category theory and afterwards, I've been approached by someone in the audience who said that talk like that made him wonder if mathematics exists at all (I'm paraphrasing here). As someone who might be labeled a formalist (which seems to be a very natural viewpoint in modern day set theory), I don't quite understand the kind of significance, some mathematicians seem to look for in the way sets are handled. $\endgroup$ – Stefan Mesken Jun 27 '16 at 10:56
  • $\begingroup$ From where I stand there is no true set of reals. We can always code something as a real and produce a larger universe that recognizes this set as a real number. What we do, in the case of real numbers and similar mathematical objects, is to prove (in a given theory $T$) that a set with certain properties exists and that this set is unique up to isomorphism - sometimes up to a unique isomorphism. Now, in any given models there is a set that has this properties. However, varying our models, there are vastly different sets that fit their model's description, having virtually nothing in common. $\endgroup$ – Stefan Mesken Jun 27 '16 at 10:59
  • $\begingroup$ @Stefan I'm a bit of a formalist too, but since material set theory doesn't jibe with how I think about sets, these sorts of issues come up. I don't really agree with the whole coding thing; what we do, in my opinion, is pin down $\mathbb{R}$ up to isomorphism. I agree that, in some sense, there's no true set of reals, because from an external perspective, the object that you thought was $\mathbb{R}$ may not be a complete ordered field after all. So there's really no reason to think there's one true $\mathbb{R}$. $\endgroup$ – goblin Jun 27 '16 at 11:00
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Let me give my take on it, set theory is the foundation of mathematics, we have sets we denote in various ways $\emptyset$ is the first from which we can build more, we have the infinite set $\omega$, from these we can make other sets with powersets, squaring, union, intersection etc. These contain elements which we can label anyway we want, $1,2,3,\ldots$, $a,b,c,\ldots$, green, blue, red etc. Those are all just labels we put on objects within the sets. They have no intrinsic values or even properties attatched to them. They are just objects we can deal with.

Now we construct natural numbers for $\mathbb{N}$ and it's cardinality is equivalent to our $\omega$. However in $\mathbb{N}$ we have now assigned elements properties and relations that $\omega$'s elements lack. When we construct the next step, integers, the question is more that we want to USE the properties of the for mentioned set, $\mathbb{N}$, to established the properties of the elements within our new set $\mathbb{Z}$. However $\mathbb{Z}$ still has the same cardinality as $\omega$, so what does that mean? All it means is that we've assigned the labels of all elements in $\omega$ differently. But it's still really the same underlying set of $\omega$ with the addition of we have a known manner to label these. In most contexts this also usually means we start invoking the properties the elements are supposed to have that we derived earlier.

This continue on when we do rational numbers, a new reassignment of $\omega$'s elements and using a previous structure to start assigning properties to these elements, but still only $\omega$ as the underlying set. For real numbers we do the same except now it's $\mathcal{P}(\omega)$ we're working with as the underlying set and relabel those and start assigning properties.

The set already existed from set theory, all you do when you construct number systems is that you assign relations, operations, properties and such to the element and the set, the underlying set however existed before.

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