# Is there a standard symbol for a multiset version of the natural numbers

In other words, is there a standard symbol for the multiset $N$ where:

$$N = \mathbb{N} \cup \mathbb{N} \cup \mathbb{N} \cup \ldots$$

Or better yet, is there another, more concise way to represent what $N$ is?

Edit:

I realize where I was going wrong. I was thinking of $\cup$ as an addition-like operator (it was late last night). I meant to define $N$ as:

$$N = \{\{1, 1, 1, \ldots, 2, 2, 2, \ldots, 3, 3, 3, \dots, \cdots\}\}$$

Edit 2:

$$N = \mathbb{N} + \mathbb{N} + \mathbb{N} + \ldots$$

Edit 3:

Here's the context I'm using it in:

Let $D_1, D_2, \ldots, D_n$ each be subsets of the multiset $N$ where $N$ is ...

As an example, $D_1$ could be $\{\{1,1,2,4,4,5\}\}$ but not $\{\{\ \frac{1}{2}, \pi, \pi, e^7\}\}$. After that, I'll never mention $N$ again (as far as I know). If there isn't a simple way to represent $N$, is it concise and clear enough to say that "every element of $D_i$ is in $\mathbb{N}$, for all $i$"?

• What does $\cup$ mean here exactly? – MJD Dec 11 '16 at 3:24
• The union of two sets. – Noah May Dec 11 '16 at 3:27
• But the union of two sets is a set, not a multiset. And if $\cup$ is set union then $\Bbb N\cup\Bbb N\cup\cdots$ is equal to $\Bbb N$. – MJD Dec 11 '16 at 13:00
• I have seen the double set brackets for multisets. Also, at least from the "context" you are trying to use it, can you just skip defining this $N$ object and say that each $D_i$ is a multiset of natural numbers? This seems to remove a lot of this headache while being very clear(in my opinion) as to what you would be talking about. – Sean English Dec 12 '16 at 16:46
• Oh, and I should mention that in many formulations, multisets do not allow for infinite instances of the same element, so depending on how exactly a multiset is defined, $N$ may not be one.(For example a common definition of multiset is an ordered pair (A,f) Where A is a set and f is a function from A to $\mathbb{N}$. Since $\infty\not\in\mathbb{N}$, it isn't possible to have infinitely many instances of any element in a multiset under this definition. – Sean English Dec 12 '16 at 16:52

First, I guess you mean that $N$ is the disjoint union of "some" copies of $\mathbb{N}$. If "some" is an integer $k$, you could represent $N$ as the set $$N = \{(i, j) \mid 1 \leqslant i \leqslant k, j \in \mathbb{N} \}$$ More generally, a disjoint union of a family of sets $(S_i)_{i \in I}$, where each $S_i$ is equal to $\mathbb{N}$, can be represented by the set $$N = \{(i, j) \mid i \in I, j \in \mathbb{N} \}$$ In this representation, $S_i = \{(i, j) \mid j \in \mathbb{N} \} = \{i\} \times \mathbb{N}$.

• Not quite. Look at my edit. – Noah May Dec 11 '16 at 20:50

You mean just a function $f: \mathbb N \rightarrow \mathbb N$? For example $f(5)$ would be the number of copies of $5$ that $f$ has. Or have the range be $\{0\} \cup \mathbb N$. Of course $f \subset \mathbb N \times \mathbb N$ as a relation, additionally with the function property.

Your first edit: Since $\{a,a\} = a$ always, you just have $N = \{\mathbb N\}$, i.e. $N$ is the set with one element $a$, and $a$ is the set with elements $1,2,3,\ldots$.

Your second edit: $N = \emptyset$ because there is no natural number that can be written as $\sum_{i=1}^\infty a_i$ where each $a_i \in \mathbb N$.

Your third edit: $D_1 = \{\{1,2,4,5\}\}$. Of course $|D_1| = 1$ and the size of the only element of $D_1$ is $4$.

• Would it be valid to say something similar to: $S \subset f$ where $S = \{\{1, 2, 2, 2, 3, 3\}\}$ and $f(1) = f(2) = f(3) = \ldots = \infty$? – Noah May Dec 11 '16 at 23:17
• @NoahMay you can't have it both ways. If you want to use standard mathematical notation then $\{a,a\} = \{a\}$, always. If you want to invent some random new notation then you have one of two choices: either you can make a new set theory that can serve as the foundation of most of mathematics, or you can find a construction in the usual set theory and notation that represents your notation. Of course, in this case, your notation should not conflict with existing, widely used notation. Which is what you keep doing when you write the same number twice and expect it to have a different meaning. – djechlin Dec 12 '16 at 16:30

As I stated in my question, the context I'm using $N$ in does not actually depend on a multiset "version" of the natural numbers. It suffices to say that "$D_i$ is a multiset of natural numbers" for all $i$, thanks to Sean English.

To answer to original question. No, there is no standard "symbol" for a multiset version of natural numbers.

• I apologize for the confusion in my statements or questions. I read an introduction to combinatorics which briefly mentioned multisets without giving a formal definition of them. So, I thought what I was saying should make sense. I realized that I was wrong with the help of G. H. Faust, dxiv, and others. – Noah May Dec 14 '16 at 19:55