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When an algebraic structure is defined, it is often defined as a set $S$ "along with"/"together with"/"having" operations $\circ_1, \circ_2, \ldots, \circ_n$, and "denoted" by $(S, \circ_1, \circ_2, \ldots, \circ_n)$.

This is true of metric spaces as well. We often say that a metric space is a set $X$ with some distance function $f: X \times X \to \mathbb{R}$.

But none of this seems very formal to me. Are we -- the reader -- supposed to infer that it in each of these cases, we're really talking about an $n$-tuple (in ZFC)?

So for instance, when a theorem says something like this:

Thm Let $X$ be a metric space. If $X$ is a foo, then $X$ is a noo.

It is formally:

Thm Let $(X, d)$ be a metric space. If (the underlying set) $X$ is a foo, then $X$ is a noo.

I guess I'm just confused as to why tuples are glossed over and replaced with "together with" so often, when it detracts from the "axiomaticity" of the material.

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    $\begingroup$ @AmadeusDrZaius At some point everyone knows that 'together with' means. I think it is a bad idea to use such terms for a beginning student. What does 'together with' even mean? Yes, you're supposed to induce (rather than infer) that an $n$-tuple is at hand. This has little to do with $\sf ZFC$ other than the fact that it is the standard foundational ground for mathematics nowadays. $\endgroup$
    – Git Gud
    Mar 26, 2014 at 18:00
  • $\begingroup$ In most cases, the reader does not have to know, nor care, whether it is an $n$-tuple in any particular way. Only a model theorist, when she wants to make this something for her to study, should specify what it is at that nuts-and-bolts level. $\endgroup$
    – GEdgar
    Mar 26, 2014 at 18:09
  • $\begingroup$ Related. $\endgroup$
    – Git Gud
    Jun 11, 2014 at 22:10

2 Answers 2

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It's often okay to define a mathematical object without an explicit set theoretic model. If we need a model we can always take a step back and make sure there is one.

Take for example the following definition:

A map $f$ from a set $A$ to a set $B$ assigns to every $a\in A$ a unique value $f(a)\in B$. We write $f\colon A\to B$. Two maps $f,g\colon A\to B$ are equal if and only if $f(a)=g(a)$ for all $a\in A$.

We can use this definition of a map to define surjectivity, injectivity and any other property of maps and work with those without ever thinking about how we would model this in set theory.

Or do you always think of a map $f\colon A\to B$ as a tuple $(F, A, B)$ where $F\subseteq A\times B$ is a right-unique relation, and the $3$-tuple $(F, A, B)$ is really build from Kuratowski pairs $(F, (A, B))$ with $(A, B) = \{\{A\}, \{A, B\}\}$?

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    $\begingroup$ You need to explain what 'assign' means if you're avoiding the set theoretic definition. $\endgroup$
    – Git Gud
    Mar 26, 2014 at 18:10
  • $\begingroup$ It's true that you don't necessarily need to build up from ZFC [one can take "assign" as an atom of our language], and you make a good point by noting that, but I think it's fair to say that much mathematical sweat and tears has been devoted to being able to put everything on a common axiomatic foundation.. so not doing so is an option, but not the option I'm looking for. $\endgroup$
    – AmadeusDrZaius
    Mar 26, 2014 at 18:10
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    $\begingroup$ @GitGud Yes, if somebody doesn't know the word, I could explain it (without set theory). I agree that modern mathematics should define every term used in mathematical statements (or at least be able to, if asked for), but to start understanding maps we certainly don't need relations and Kuratowski pairs. $\endgroup$
    – Christoph
    Mar 26, 2014 at 18:18
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    $\begingroup$ Defining 'assigning' without set theory is simply shifting the problem to another foundational ground. I agree that in the case of functions, for starters, it is best to use something like what you suggested, but this is for purely pedagogical reasons and that's not what's at stake in this question. In regards to 'together with' I think not defining as an $n$-tuple just brings confusion. I know I was confused and found relief in learning the $n$-tuple definition. $\endgroup$
    – Git Gud
    Mar 26, 2014 at 18:33
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    $\begingroup$ I agree. Though I have never seen a rigorous definition of a category being a $3$-tuple of two classes and a relation in some theory where tuples can hold classes. And I never felt the need to. $\endgroup$
    – Christoph
    Mar 26, 2014 at 18:40
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One can, but I do not think one should.

Also, $n$-tuples are not basic objects of standard set theories, they have to be defined. Take a probability space $(\Omega,\Sigma,\mu)$. Should we take this as $(\Omega,(\Sigma,\nu))$ with some definition of an ordered pair? How about $((\Omega,\Sigma),\mu)$? Or maybe it is a function $f$ with domain $\{0,1,2\}$ such that $f(0)=\Omega$, $f(1)=\Sigma$, $f(2)=\nu$. ? Many other choices are possible, but it doesn't really matter. If a result in, say, geometry essentially depends on how you define $n$-tuples it is probably of no interest to any geometer.

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    $\begingroup$ Then you need to explain what 'together with' means. $\endgroup$
    – Git Gud
    Mar 26, 2014 at 18:08
  • $\begingroup$ Well, we have to give it some definition. Also, if you define $(A, B, C)$ as $\{\{A\}, \{A, B\}, \{A, B, C\}\}$, and so on for other $n$, then there is no ambiguity.. $\endgroup$
    – AmadeusDrZaius
    Mar 26, 2014 at 18:08
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    $\begingroup$ The point isn't that 'a 3-tuple' is ambiguous, it's that a geometer doesn't care about the ambiguity. $\endgroup$
    – user98602
    Mar 26, 2014 at 21:27

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