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

Question

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 $i$"?

Answer 1

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.

Answer 2

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, $S_i = \{(i, j) \mid j \in \mathbb{N} \} = \{i\} \times \mathbb{N}$.

Answer 3

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 $4$.