Originally I thought that in statistics, a data set is just a set of real numbers, and that was it. But in the case of a set, there can only be one instance of any given entry, e.g. in set theory

$$\left \{ 1,2,2 \right \}=\left \{ 1,2 \right \}$$

A set of this form in set theory may also be called an unordered pair.

But from the point of view of statistics the objects on both sides of the equation are distinct, e.g. the left one has a mode, while the right one doesn't; the left one's arithmetic mean is $\frac{5}{3}$, while the arithmetic mean of the right one is $\frac{3}{2}$.

Question: are the concepts of data set in statistics and set in set theory really different things? Are data sets of real numbers in statistics really $n$-tuples of set theory in disguise, or taken to be ones most of the time implicitly?

  • 1
    $\begingroup$ to get order from unordered sets you can always just nest things like so $\{1,\{2\},\{\{2\}\}\}$, or using some other scheme. If you decide you really don't want an explicit ordering, take a ''dataset'' to be an equivalence class of the relation on ordered sets in the above sense that makes permutations equivalent. $\endgroup$ – enthdegree Jul 25 '16 at 11:20
  • 7
    $\begingroup$ You also have the concept called a multiset, which pairs each value with how many times it's been observed, so you would actually have the unordered set $\{(1, 1), (2,2)\}$ of ordered pairs, saying that $1$ was observed once, and $2$ was obseved twice. This does not introduce any ordering you later ignore, and it makes the process of fetching a value consistent. $\endgroup$ – Arthur Jul 25 '16 at 11:21
  • 3
    $\begingroup$ Why is this downvoted? It would be nice if the downvoters explain their decision. $\endgroup$ – Björn Friedrich Jul 25 '16 at 11:33
  • 1
    $\begingroup$ related: math.stackexchange.com/questions/18024/set-vs-multiset see also discussion in wikipedia en.wikipedia.org/wiki/Talk%3AData_set $\endgroup$ – leonbloy Jul 25 '16 at 13:06
  • 1
    $\begingroup$ @BjörnFriedrich It was probably downvoted because it's about word semantics. I'd dowvote "is a strange quark actually strange?" in the Physics SE. For that matter, is a television set really a set? The components have to be arranged into a specific graph in order to receive, decode and render the video and audio. It should be called a television graph! $\endgroup$ – Kaz Jul 26 '16 at 14:17

Arthur is right; the term "data set" usually means multiset. For example, a bivariate data set just means a multiset of elements of $\mathbb{R}^2$. Furthermore, if $X$ is a set, I like to write $\mathbb{N}\langle X \rangle$ for the set of all multisets in $X$. Therefore, $\mathbb{N}\langle \mathbb{R}^2\rangle$ is notation for the collection of all bivariate data sets. The remainder of my answer will address the question:

What is a multiset?

Informally, a multiset in $X$ is like a finite subset of $X$, except that repetitions are allowed. (Order still doesn't matter.) For example, the following are multisets in $\mathbb{N}$: $$\{1,2\} \quad \{2,1\} \quad \{2,1,1\}$$ The first two are equal, but the last one is distinct from the other two. Its sometimes clearer to write multisets using the notation of linear combinations: $$\{a,b,b\} = \{a\}+\{b\}+\{b\} = \{a\}+2\{b\}$$

Here's a few different formalizations, starting from the most concrete and ending with the most abstract:

Let $X$ denote a set. Then:

Definition 0. A multiset in $X$ is a finitely-supported function $X \rightarrow \mathbb{N}$.

(The support of $f : X \rightarrow \mathbb{N}$ is defined to be $\{x \in X \mid f(x) \neq 0\}$, and $f$ is said to be finitely-supported iff this set is finite.)

More abstractly:

Definition 1. A multiset in $X$ is an element of the $\mathbb{N}$-module freely generated by $X$.

(This explains why the notation of linear-combinations works. It also explains why $\mathbb{N}\langle X\rangle$ is good notation.)

To see how and why Definition 0 works, just interpret $\{a\}$ as a function $X \rightarrow \mathbb{N}$ for each $a \in X$ as follows: $$\{a\}(b) = [a=b]$$

(See also, Iverson bracket. I usually avoid Kronecker delta notation because its a less a versatile formalization and you can just do a lot less with it; therefore, I think the mathematical community should phase out its usage.)

Now observe that the set of finitely-supported functions $X \rightarrow \mathbb{N}$ form an $\mathbb{N}$-module under the pointwise operations. This allows us to add up elements of the form $\{a\}$ however we please, essentially building complicated multisets from simpler "atoms." We can say more: the set $$\{\{a\} : a \in X\}$$ is a basis for the set of all finitely-supported functions $X \rightarrow \mathbb{N}$, which explains the equivalence with Definition 1. In fact, it is the only basis.

For the more advanced reader:

First, some comments of a general nature. If you've only considered modules only over rings, the uniqueness of a basis might come as a bit of a shock. This is all possible because $\mathbb{N}$ is not a ring, of course. Another case of this is $\mathbb{B}$-modules, where $\mathbb{B} = \{0,1\}$ has multiplication given by Logical AND, addition given by Logical OR. So in particular, $1+1 = 1$, in contrast to the ring $\mathbb{Z}/2\mathbb{Z}$, which has $1+1 = 0$. Anyway, a $\mathbb{B}$-module turns out to be the same thing as a unital semilattice, and the $\mathbb{B}$-module freely by $X$ is just $\mathcal{P}_{\mathrm{fin}}(X)$, the collection of all finite subsets of $X$. The singletons provide the unique basis.

Moving beyond semirings, another place where basis-uniqueness occurs is in the context of barycentric algebras. In this case, the free algebras are simplices, which explains how we're able to speak of the vertexes of a simplex.

On another topic altogether, we can also try to categorify:

Definition 2. A categorified multiset in $X$ is a finitely-supported function $X \rightarrow \mathbf{FinSet}$.

(In other words, a categorified multiset is an $X$-indexed family of finite sets such that all but finitely many of those sets are empty.)

More abstractly:

Definition 3. A categorified multiset in $X$ is an object of the finite-coproduct category freely generated by $X$.

I'd also add that there's definition that seems not to fit the above scheme:

Definition. 4 A categorified multiset in $X$ is an object $M$ of the slice category $\mathbf{Set}/X$ such that the underlying set of $M$ is finite.

  • $\begingroup$ Please, explain what a multiset is in the answer's body so that people don't have to look through the comments. $\endgroup$ – Michael Smith Jul 25 '16 at 11:38
  • 2
    $\begingroup$ @MichaelSmith, just did the crap out of that request :) $\endgroup$ – goblin Jul 25 '16 at 11:45
  • 1
    $\begingroup$ I find the edit really funny and witty, I really do. But I'm really worried about some people not getting the main idea at all. Please, reedit the question framing it in terms of one definition involving Cartesian products (as @Arthur has already done). Thank you very much, my friend :) $\endgroup$ – Michael Smith Jul 25 '16 at 12:44
  • 1
    $\begingroup$ @MichaelSmith, how's that? I should add that personally, I'm quite partial to Definition 3, because I think it probably generalizes the best out of all of them. Let $\mathbf{G}$ denote a groupoid. Then a multiset in $\mathbf{G}$ can be defined as an object of the finite-coproduct completion of $\mathbf{G}$. I suspect that this is probably the correct categorification of this concept. $\endgroup$ – goblin Jul 25 '16 at 14:23

A "data set" in statistics does indeed allow repetitions and in that sense is different from a "set" in set theory.

It wouldn't make much sense otherwise: for instance, if you take the average daily temperature of each day for a year, there are only going to be a couple of dozen values (or a few dozen, in Fahrenheit), and the concept of average or mean, standard deviation, and so on, wouldn't make any sense.

Depending on the context, a "data set" is either an ordered series of values (thus, an $n$-tuple in disguise, as you say) or a collection of values, some of which may be repeated, with no ordering implied - so that $\{1,2,2\} = \{2,1,2\} = \{2,2,1\}$. I suppose that if you were desperate then you could consider the latter as a map from the space of possible values to the set of natural numbers.


In most cases the data set will be a true set if you view it as a set of observations. In the example of temperatures, there are only a few different temperatures, but each one corresponds to a different day. Your data set consists of ordered pairs (day, temperature on that day) and no pair is repeated. The only way you get repetition is if you observe the same data more than once. If you have a data set of the number of legs on horses, your observations are (horse name, number of legs). If you have a repeat, you have observed the same horse more than once, so you might want to delete one observation. Alternately, you might worry that a horse had lost a leg, in which case your data is (horse name, date observed, number of legs) and again you won't have any duplicates.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.