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I'm not a set theorist, but I feel like I've been taught all my life that I can write sets with repeating elements but those repeats are sort of degenerate and not counted as separate. So $\{a, b, a\} = \{a, b\}$. This always fit nicely with the fact that I could only check if two sets are equal by asking if they are subsets of one another. But a (computer science with math background) colleague of mine got into a long debate with me on whether I'm ever allowed to do this without referring to some multiset. For a moment I said, OK, maybe I've just not been careful. But now I'm seeing this issue pop up everywhere. Below is a recent example from notes I was writing for my students.

Can anyone help me with a reference to the fact that $\{a, b, a\} = \{a, b\}$ is definitely OK or not OK??

Example: For a group $G$ and an element $a \in G$, define $$\langle a \rangle := \{ a^n \in G \vert n \in \mathbb{Z} \}$$

Of course if the group is finite then for some $n \ne m$ we have $a^n = a^m$ and I feel like this is used a lot to describe this fairly reasonable set. In some sense it has repeating elements but if you take the approach above it's not an issue at all. What am I missing?

cheyne
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    In situations where you want them to mean something different, you can specify that you want them to mean something different. In situations where you want them to mean the same thing, you can specify that you want them to mean the same thing. That's the thing about math... we get to choose the rules we work with and definitions we follow, and sometimes there are some people who prefer one way and others who prefer another. – JMoravitz Feb 24 '23 at 13:25
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    As for an explicit reference... surely every introductory text on set theory will say exactly what you are suggesting. Elsewhere here on this site as well... such as here. I would say your motivating example is also perfectly good at conveying a situation where although we expect there to be duplicates we don't mind that at all. Consider also a simpler one of ${a}\cup {b}={a,b}$ which we could still write in this way without the need to explicitly state the caveat that if $a=b$ that the above can be simplified. – JMoravitz Feb 24 '23 at 13:31
  • Thanks for the comments @JMoravitz. The link you included has this very same argument in the comments. And no clear references. I agree with you. I'm an algebraic topologist who learned set theory from Munkres' "Topology" and never looked back. I'm currently looking in Halmos' Naive Set Theory as suggested in that linked answer and .... not quite found it.

    See if I had written in my question ${x \in G \vert x = a^n \text{ for some } n \in \mathbb{Z} }$ I would be avoiding this entirely. Which is what Halmos does and has made me question lots of things.

     – cheyne Feb 24 '23 at 13:42
  • In a set, repetitions of elements don't matter. In a multiset, they do. Equivalently: Two sets are equal if they have the same elements (i.e., every element of either is also an element of the other). Two multisets are equal if they have the same elements with the same multiplicities. – Andreas Blass Feb 25 '23 at 00:20
  • @AndreasBlass I agree with you and have for 20 years. I don't know the reference. – cheyne Feb 25 '23 at 23:27
  • I'm going to see what others chime in on before accepting other answers, but Michael's answer is what I was hoping for so I'm accepting that immediately. – cheyne Feb 25 '23 at 23:31

3 Answers3

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I believe the best reference would include the reason why duplicate elements are extraneous.

It's due to the Axiom of Extentionality.

From Kunen's, Set Theory: An Introduction to Independence Proofs

Axiom of Extentionality

$$\forall x\forall y[\forall z(z\in x\iff z\in y) \iff x=y]$$

In English, two sets are equal if and only if they have the same elements.

So, $\{a,b,c\}= \{a,a,b,b,c,c\}$ by the Axiom.

On page 12, you see the explicit example that $\{x,x\} = \{x\}$.

To see why this generalizes (from a different perspective)

Notice that by the Axiom of Extentionality,

$$\forall x(z\in x\Rightarrow x\cup\{z\} =x)$$

So, adding elements to a set, that are already in it- doesn't change the set.

Michael Carey
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  • Lovely! Thank you @michael. I appreciate the reference along with the "On page 12...." that was so helpful. – cheyne Feb 25 '23 at 23:28
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Michael Carey has explained why, on classical set theory, $\{a,b,a\}= \{a,b\}$. It is due to the axiom of extensionnality.

Now, if you want, you can give a sense to the notion of "multiset" in classical set theory. Basically $\{a,b,a\}$ is the set $\{a,b\}$ where you count $a$ $2$ times and $b$ $1$ time. So we will say, by definition, that $\{a,b,a\}$ is a function $f : \{a,b\} \to \mathbb{N}$ such that $f(a)=2$ and $f(b)=1$.

More generally, starting from a set $S$, you can define all the multisets generated by $S$ as all the possible functions $f : S \to \mathbb{N}.$

C. Dubussy
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let's call the classification proposed by CS colleague a two-fold way : there are sets, and there are multi-sets.

The most known classification that places sets and multi-sets in a larger context is the so called Twelve-fold Rota way : https://en.wikipedia.org/wiki/Twelvefold_way#Intuitive_meaning_of_the_chart_using_Balls_and_Boxes_scenario

There is also a Bogart twenty fold way https://en.wikipedia.org/wiki/Twelvefold_way#The_twentyfold_way

In introductory texbooks there is also a simple classification, a 4-fold way, that consider only lists and sets, with or without repetitions.

A 30-fold way of placing in context sets and multisets is here (extending Rota way) https://arxiv.org/ftp/math/papers/0606/0606404.pdf (this one can be completed to a 40-fold ways).

I would ask the C.S. collegue "how many empty sets do you consider in C.S. ?" given that in Combinatorics invoked here we have different empty boxes, any amount, for free, not defined by extensionality but by color, position or other qualities (intensionality).

Boyku
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