# Results on Functions on Dedekind Multisets

^{1}

^{2}

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## Abstract

**:**

## 1. Introduction

**f**initely

**r**epeated

**e**lement

**s**et), heap, bunches, etc. These concepts have all been studied by various mathematicians with different applications.

## 2. Preliminaries

**Definition**

**1.**

**Definition**

**2.**

**Definition**

**3.**

**Definition**

**4.**

- i.
- Consider $A\in MS\left(X\right)$. The insertion of x into A results into a multiset denoted by $\mathcal{C}=x\uplus A$ which has the count function$${C}_{\mathcal{C}}\left(y\right)=\left\{\begin{array}{c}{C}_{A}\left(y\right),\phantom{\rule{2.em}{0ex}}y\ne \phantom{\rule{3.33333pt}{0ex}}x\hfill \\ \phantom{\rule{4pt}{0ex}}\hfill \\ {C}_{A}\left(x\right)+1,\phantom{\rule{0.277778em}{0ex}}y=x.\hfill \end{array}\right.$$
- ii.
- Consider $A,B\in MS\left(X\right)$. The insertion of A into B or of B into A results into a multiset $\mathcal{C}$ which has the count function denoted by ${C}_{\mathcal{C}}\left(x\right)={C}_{A}\left(x\right)+{C}_{B}\left(x\right)$.

**Definition**

**5.**

- i.
- Consider $A\in MS\left(X\right)$. The removal of x from A results into a multiset denoted by $\mathcal{D}=A\ominus x$ which has the count function$${C}_{\mathcal{D}}\left(y\right)=\left\{\begin{array}{c}max\{{C}_{A}\left(y\right)-1,0\},\phantom{\rule{0.277778em}{0ex}}y=x\hfill \\ \phantom{\rule{4pt}{0ex}}\hfill \\ {C}_{A}\left(y\right),\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}y\ne x.\hfill \end{array}\right.$$
- ii.
- Consider $A,B\in MS\left(X\right)$. The removal of B from A results into a multiset which has the count function denoted by ${C}_{\mathcal{D}}\left(x\right)=max\{{C}_{A}\left(x\right)-{C}_{B}\left(x\right),0\}.$

**Definition**

**6.**

- i.
- The multiset $\mathcal{E}=A\otimes B$ is such that $\mathcal{E}$ only contains elements of A which also occur in B. The count function of $\mathcal{E}$ is denoted by$${C}_{\mathcal{E}}\left(x\right)=\left\{\begin{array}{c}{C}_{A}\left(x\right),\phantom{\rule{0.277778em}{0ex}}x\in \phantom{\rule{3.33333pt}{0ex}}B\hfill \\ \phantom{\rule{4pt}{0ex}}\hfill \\ 0,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}x\notin B.\hfill \end{array}\right.$$
- ii.
- The multiset $\mathcal{F}=A\u2a00B$ is such that $\mathcal{F}$ only contains elements of A which do not occur in B. The count function of $\mathcal{F}$ is denoted by$${C}_{\mathcal{F}}\left(x\right)=\left\{\begin{array}{c}{C}_{A}\left(x\right),\phantom{\rule{0.277778em}{0ex}}x\notin B\hfill \\ \phantom{\rule{4pt}{0ex}}\hfill \\ 0,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}x\in B.\hfill \end{array}\right.$$

**Definition**

**7.**

- i.
- ${C}_{M}\left(xy\right)\ge {C}_{M}\left(x\right)\wedge {C}_{M}\left(y\right),$
- ii.
- ${C}_{M}\left({x}^{-1}\right)\ge {C}_{M}\left(x\right).$

**Remark**

**1.**

**Example**

**1.**

- i.
- ${C}_{N}\left(1\right)=2$, ${C}_{N}\left(2\right)=2$, ${C}_{N}\left(3\right)=1$, ${C}_{N}\left(4\right)=3$ and ${C}_{N}\left(5\right)=1$
- ii.
- $M\cap N=\{1,1,2,2,3,4,4,5\}$. When two or more multisets are intersected, the minimum multiplicity of the common elements is taken.
- iii.
- $M\cup N=\{1,1,1,1,2,2,3,3,3,4,4,4,5,5\}$. When the union of two or more multisets is taken, the maximum multiplicity of the common elements is taken.
- iv.
- Neither is $M\subseteq N$, since ${C}_{M}\left(x\right)\le {C}_{N}\left(x\right)$ for all $x\in X$, nor $M\supseteq N$, since ${C}_{M}\left(x\right)\ge {C}_{N}\left(x\right)$ for all $x\in X$, but $W\subseteq M$, since ${C}_{W}\left(x\right)\le {C}_{M}\left(x\right)$ for all $x\in X$.

**Example**

**2.**

- i.
- Let $G=\{1,-1,i,-i\}$ be a group with the usual multiplication. $M=\{1,1,1,-1,-1,-1,i,i,-i,-i\}$ is a multigroup since it satisfies Definition 7 and Remark 1.
- ii.
- Consider the group ${Z}_{3}=\{0,1,2\}$ with the modulo addition. $M=\{0,0,0,1,1,$$2,2\}$ is a multigroup.

**Definition**

**8.**

- i.
- ${A}_{n}=\{x:{C}_{A}\left(x\right)\ge n\}$;
- ii.
- We denote a multiset containing only one element x with multiplicity n as ${\left[n\right]}_{x}$–a simple multiset;
- iii.
- The complement of the multiset $M\in {\left[X\right]}^{\alpha}$ denoted by ${M}^{\prime}$ is such that ${C}_{{M}^{\prime}}\left(x\right)=\alpha -{C}_{M}\left(x\right)$;
- iv.
- $nA=\{{x}^{n},\phantom{\rule{0.277778em}{0ex}}\forall x\in A\phantom{\rule{0.277778em}{0ex}},n\phantom{\rule{0.277778em}{0ex}}is\phantom{\rule{0.277778em}{0ex}}the\phantom{\rule{0.277778em}{0ex}}multiplicity\phantom{\rule{0.277778em}{0ex}}of\phantom{\rule{0.277778em}{0ex}}each\phantom{\rule{0.277778em}{0ex}}element\phantom{\rule{0.277778em}{0ex}}that\phantom{\rule{0.277778em}{0ex}}appears\phantom{\rule{0.277778em}{0ex}}in\phantom{\rule{0.277778em}{0ex}}nA\}.$

**Example**

**3.**

- i.
- $2A=\{0,0,0,0,0,0,1,1,1,1,2,2,2,2\}$, the multiplicities of each element in A is doubled;
- ii.
- ${\left[2\right]}_{0}=\{0,0\}$, this multiset comprises only 0 and its multiplicity is 2;
- iii.
- ${A}_{2}={A}_{1}={A}_{0}=\{0,1,2\}$ and ${A}_{3}=\left\{0\right\}.$ These are the sets of elements of A respectively with multiplicities 2, 1, 0 and 3.

**Remark**

**2.**

**Proposition**

**1.**

- i.
- If $A\subseteq B$, then ${A}_{n}\subseteq {B}_{n}$;
- ii.
- If $m\le n$, then ${A}_{m}\supseteq {A}_{n}$;
- iii.
- ${(A\cap B)}_{n}={A}_{n}\cap {B}_{n}$;
- iv.
- ${(A\cup B)}_{n}={A}_{n}\cup {B}_{n}$;
- v.
- $A=B$ if and only if ${A}_{n}={B}_{n},\phantom{\rule{0.277778em}{0ex}}\forall n\in I\phantom{\rule{-0.166667em}{0ex}}\phantom{\rule{-0.166667em}{0ex}}N$.

**Definition**

**9.**

- i.
- the image of M under f denoted $f\left(M\right)$ has the count function$${C}_{f\left(M\right)}\left(y\right)=\left\{\begin{array}{c}\underset{f\left(x\right)=y}{\bigvee}{C}_{M}\left(x\right),\phantom{\rule{0.277778em}{0ex}}if\phantom{\rule{0.277778em}{0ex}}{f}^{-1}\left(y\right)\ne \xd8\hfill \\ \phantom{\rule{4pt}{0ex}}\hfill \\ 0,\phantom{\rule{2.em}{0ex}}\phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}otherwise;\hfill \end{array}\right.$$
- ii.
- the inverse image of N under f denoted ${f}^{-1}\left(N\right)$ has the count function ${C}_{{f}^{-1}\left(N\right)}\left(x\right)={C}_{N}\left[f\left(x\right)\right]$.

**Proposition**

**2.**

- i.
- ${M}_{1}\subseteq {M}_{2}\Rightarrow f\left({M}_{1}\right)\subseteq f\left({M}_{2}\right)$;
- ii.
- $f\left({\cup}_{i\in I}{M}_{i}\right)={\cup}_{i\in I}f\left({M}_{i}\right)$;
- iii.
- ${N}_{1}\subseteq {N}_{2}\Rightarrow {f}^{-1}\left({N}_{1}\right)\subseteq {f}^{-1}\left({N}_{2}\right)$;
- iv.
- ${f}^{-1}\left({\cup}_{i\in I}{M}_{i}\right)={\cup}_{i\in I}{f}^{-1}\left({M}_{i}\right)$;
- v.
- ${f}^{-1}\left({\cap}_{i\in I}{M}_{i}\right)={\cap}_{i\in I}{f}^{-1}\left({M}_{i}\right)$;
- vi.
- $f\left({M}_{i}\right)\subseteq {N}_{j}\Rightarrow {M}_{j}\subseteq {f}^{-1}\left({N}_{j}\right)$;
- vii.
- $g\left[f\left({M}_{i}\right)\right]=\left[gf\right]\left({M}_{i}\right)$ and ${f}^{-1}\left[{g}^{-1}\left({N}_{j}\right)\right]={\left[gf\right]}^{-1}\left({N}_{j}\right)$.

## 3. Some Illustrations of Properties of Operations on Multisets

**Proposition**

**3.**

- i.
- Let A be a multiset drawn a nonempty set X. The n insertion of A into itself denoted ${\uplus}_{n}A=nA$;
- ii.
- $A\uplus A=A\iff A=\varphi $.

**Proof.**

- By Definition 4(ii),$${C}_{{\uplus}_{n}A}\left(x\right)=\underset{n\phantom{\rule{0.277778em}{0ex}}times}{\underbrace{{C}_{A}\left(x\right)+{C}_{A}\left(x\right)+\cdots +{C}_{A}\left(x\right)}}=n{C}_{A}\left(x\right)\phantom{\rule{0.277778em}{0ex}}\forall x\in A;$$
- Assume that $A\uplus A=A$ but $A\ne \xd8$. By (i), $A\uplus A=2A$. Then, $2A=A$. This is not possible by Definition 4(ii). Thus, $A=\xd8$.Conversely, assume that $A=\xd8$ but $A\uplus A\ne A=\xd8$. Then $A\uplus A=2A$. But ${C}_{A}\left(x\right)=0;\phantom{\rule{0.277778em}{0ex}}\forall x\in X.$ Then, ${C}_{A\uplus A}\left(x\right)={C}_{A}\left(x\right)+{C}_{A}\left(x\right)=0$. This implies that $A\uplus A=\xd8$. This is a contradiction. Hence, $A\uplus A=A.$

**Remark**

**3.**

- i.
- Let $D=A\u2a02B$. If $B\subset {A}^{*}$ then $D\subseteq A$. If ${A}^{*}\subseteq B$ then $D=A$.
- ii.
- If $B\subset {A}^{*}$ then $E\subseteq A$. If ${A}^{*}\subseteq B$ then $E=\xd8$.
- iii.
- ${D}^{*}\cup {E}^{*}={A}^{*}$.

**Example**

**4.**

- i.
- If $A=\{1,1,1,2,3,3,3,4,4,5,5\}$. Then, $D=\{2,3,3,3,5,5\}$, $E=\{1,1,1,4,4\}$ and ${A}^{*}=\{1,2,3,4,5\}$. Obviously, $B\subseteq {A}^{*}$, $D\subseteq A$ and $E\subseteq A$.
- ii.
- If, on the other hand, $A=\{3,3,3,5,5\}$, ${A}^{*}=\{3,5\}$. Then, ${A}^{*}\subseteq B$, $D=\{3,3,3,5,5\}=A$ and $E=\xd8$.
- iii.
- In [i], ${D}^{*}=\{2,3,5\}$ and ${E}^{*}=\{1,4\}$. Hence, ${D}^{*}\cup {E}^{*}={A}^{*}$. Also, in [ii] ${D}^{*}=\{3,5\}$ and ${E}^{*}=\xd8$. Hence, ${D}^{*}\cup {E}^{*}={A}^{*}$.

**Proposition**

**4.**

**Proof.**

**Proposition**

**5.**

- i.
- $A\cup B\subseteq A\uplus B$;
- ii.
- $A\cup B=A\uplus B$ if $A\cap B=\xd8$.

**Proof.**

- ${C}_{A\cup B}\left(x\right)={C}_{A}\left(x\right)\vee {C}_{B}\left(x\right)=max(n,m)\le n+m={C}_{A}\left(x\right)+{C}_{B}\left(x\right)={C}_{A\uplus B}\left(x\right)$;
- Since $A\cap B=\xd8$, if $x\in A$, ${C}_{A}\left(x\right)=n$ and ${C}_{B}\left(x\right)=0$. On the other hand, if $x\in B$, ${C}_{B}\left(x\right)=m$ and ${C}_{A}\left(x\right)=0$. ${C}_{A\cup B}\left(x\right)={C}_{A}\left(x\right)\vee {C}_{B}\left(x\right)=max(n,m)=n+m={C}_{A}\left(x\right)+{C}_{B}\left(x\right)={C}_{A\uplus B}\left(x\right)$.

**Definition**

**10.**

**Proposition**

**6.**

- i.
- If $m\le n$, then ${A}_{m}^{\prime}\subseteq {A}_{n}^{\prime}$
- ii.
- $A\subseteq B\Rightarrow {B}_{n}^{\prime}\subseteq {A}_{n}^{\prime}$.
- iii.
- ${A}_{n}^{\prime}\cup {B}_{n}^{\prime}={(A\cap B)}_{n}^{\prime}$;
- iv.
- ${A}_{n}^{\prime}\cap {B}_{n}^{\prime}={(A\cup B)}_{n}^{\prime}.$

**Proof.**

- The proof is evident.
- Since $A\subseteq B,{C}_{B}\left(x\right)\ge {C}_{A}\left(x\right)$. Let $x\in {A}_{n},{C}_{A}\left(x\right)\ge n$. But ${C}_{B}\left(x\right)\ge {C}_{A}\left(x\right)\ge n$. Then $x\in {B}_{n}$ and that implies that ${A}_{n}\subseteq {B}_{n}$. From elementary set theory, ${B}_{n}^{\prime}\subseteq {A}_{n}^{\prime}$.
- Let $x\in {(A\cap B)}_{n}^{\prime}$ then ${C}_{(A\cap B)}\left(x\right)<n$. Thus, $min\{{C}_{A}\left(x\right),{C}_{B}\left(x\right)\}<n$ in which case, ${C}_{A}\left(x\right)<n$ or ${C}_{B}\left(x\right)<n$. We conclude that $x\in {A}_{n}^{\prime}$ or $x\in {B}_{n}^{\prime}$. Hence, $x\in ({A}_{n}^{\prime}\cup {B}_{n}^{\prime})$ and ${(A\cap B)}_{n}^{\prime}\subseteq ({A}_{n}^{\prime}\cup {B}_{n}^{\prime})$.Now let $x\in ({A}_{n}^{\prime}\cup {B}_{n}^{\prime})$. $x\in {A}_{n}^{\prime}$ or $x\in {B}_{n}^{\prime}$ or both. Then, $x\notin {A}_{n}$ or $x\notin {B}_{n}$ or not in both. ${C}_{A}\left(x\right)<n$ and ${C}_{B}\left(x\right)<n$. Therefore, ${C}_{A}\left(x\right)\wedge {C}_{B}\left(x\right)={C}_{(A\cap B)}\left(x\right)<n$. We conclude that $x\in {(A\cap B)}_{n}^{\prime}$ and $({A}_{n}^{\prime}\cup {B}_{n}^{\prime})\subseteq {(A\cap B)}_{n}^{\prime}$.
- Let $x\in ({A}_{n}^{\prime}\cap {B}_{n}^{\prime})$. Then, $x\in {A}_{n}^{\prime}$ and $x\in {B}_{n}^{\prime}$, which implies that $x\notin {A}_{n}$ and $x\notin {B}_{n}$. Thus, ${C}_{A}\left(x\right)<n$ and ${C}_{B}\left(x\right)<n$. As a result, ${C}_{A}\left(x\right)\vee {C}_{B}\left(x\right)={C}_{A\cup B}\left(x\right)<n$. Consequently, $x\in {(A\cup B)}_{n}^{\prime}\Rightarrow {A}_{n}^{\prime}\cap {B}_{n}^{\prime}\subseteq {(A\cup B)}_{n}^{\prime}$.On the other hand, let $x\in {(A\cup B)}_{n}^{\prime}$. Then, $x\notin {(A\cup B)}_{n}$ and ${C}_{A\cup B}\left(x\right)<n$. Furthermore, if ${C}_{A\cup B}\left(x\right)={C}_{A}\left(x\right)\vee {C}_{B}\left(x\right)<n$, then ${C}_{A}\left(x\right)<n$ and ${C}_{B}\left(x\right)<n$. The consequence is that $x\in {A}_{n}^{\prime}$ and $x\in {B}_{n}^{\prime}$. Hence, $x\in ({A}_{n}^{\prime}\cap {B}_{n}^{\prime})\Rightarrow {(A\cup B)}_{n}^{\prime}\subseteq ({A}_{n}^{\prime}\cap {B}_{n}^{\prime})$. □

## 4. Results on Function on Multisets

**Definition**

**11.**

**Example**

**5.**

**Definition**

**12.**

**Proposition**

**7.**

- i.
- ${M}_{1}\subseteq {M}_{2}\Rightarrow f\left({M}_{1}\right)\subseteq f\left({M}_{2}\right)$;
- ii.
- $f\left({\cup}_{i\in I}{M}_{i}\right)\supseteq {\cup}_{i\in I}f\left({M}_{i}\right)$;
- iii.
- ${N}_{1}\subseteq {N}_{2}\Rightarrow {f}^{-1}\left({N}_{1}\right)\subseteq {f}^{-1}\left({N}_{2}\right)$;
- iv.
- ${f}^{-1}\left({\cup}_{i\in I}{N}_{i}\right)={\cup}_{i\in I}{f}^{-1}\left({N}_{i}\right)$;
- v.
- ${f}^{-1}\left({\cap}_{i\in I}{N}_{i}\right)={\cap}_{i\in I}{f}^{-1}\left({N}_{i}\right)$;
- vi.
- $f\left({M}_{i}\right)\subseteq {N}_{j}\Rightarrow {M}_{i}\subseteq {f}^{-1}\left({N}_{j}\right)$;
- vii.
- $g\left[f\left({M}_{i}\right)\right]=\left[gf\right]\left({M}_{i}\right)$ and ${f}^{-1}\left[{g}^{-1}\left({W}_{j}\right)\right]={\left[gf\right]}^{-1}\left({W}_{j}\right)$.

**Proof.**

- Suppose ${M}_{1}\subseteq {M}_{2}$ and let $y\in f\left({M}_{1}\right)$. ${C}_{f\left({M}_{1}\right)}\left(y\right)=\sum {C}_{{M}_{1}}\left(x\right)\le \sum {C}_{{M}_{2}}\left(x\right)={C}_{f\left({M}_{2}\right)}\left(y\right)$;
- Note that ${C}_{\cup {M}_{i}}\left(x\right)\ge {C}_{{M}_{i}}\left(x\right)\Rightarrow \sum {C}_{\cup {M}_{i}}\left(x\right)\ge \sum {C}_{{M}_{i}}\left(x\right)$. Then, ${C}_{f(\cup {M}_{i})}\left(y\right)=\sum {C}_{\cup {M}_{i}}\left(x\right)=\sum \vee {C}_{{M}_{i}}\left(x\right)\ge \vee \sum {C}_{{M}_{i}}\left(x\right)=\vee {C}_{f\left({M}_{i}\right)}\left(y\right)={C}_{\cup f\left({M}_{i}\right)}\left(y\right)$;
- Let ${M}_{1}\subseteq {M}_{2}$. Then, $f\left({M}_{1}\right)={N}_{1}\subseteq {N}_{2}=f\left({M}_{2}\right)$. Let $x\in {f}^{-1}\left({N}_{1}\right)$. ${C}_{{f}^{-1}\left({N}_{1}\right)}\left(x\right)={C}_{{N}_{1}}\left(f\left(x\right)\right)=\sum {C}_{{M}_{1}}\left(x\right)\le \sum {C}_{{M}_{2}}\left(x\right)={C}_{{N}_{2}}\left(f\left(x\right)\right)={C}_{{f}^{-1}\left({N}_{2}\right)}\left(x\right)$.
- ${C}_{{f}^{-1}\cup {N}_{i}}\left(x\right)={C}_{\cup {N}_{i}}\left(f\left(x\right)\right)=\vee {C}_{{N}_{i}}\left(f\left(x\right)\right)=\vee {C}_{{f}^{-1}\left({N}_{i}\right)}\left(x\right)={C}_{\cup {f}^{-1}\left({N}_{i}\right)}\left(x\right)$.
- ${C}_{{f}^{-1}\cap {N}_{i}}\left(x\right)={C}_{\cap {N}_{i}}\left(f\left(x\right)\right)=\wedge {C}_{{N}_{i}}\left(f\left(x\right)\right)=\wedge {C}_{{f}^{-1}\left({N}_{i}\right)}\left(x\right)={C}_{\cap {f}^{-1}\left({N}_{i}\right)}\left(x\right)$.
- Let $f\left({M}_{i}\right)\subseteq {N}_{j}$ and $x\in {M}_{i}$ such that $f\left(x\right)=y$. ${C}_{{M}_{i}}\left(x\right)={C}_{f\left({M}_{i}\right)}\left(y\right)\le {C}_{{N}_{j}}\left(y\right)={C}_{{f}^{-1}\left({N}_{j}\right)}\left(x\right)$.
- ${C}_{g\left[f\left({M}_{i}\right)\right]}\left(z\right)=\sum _{{g}^{-1}\left(z\right)\ne \xd8g\left(y\right)=z}{C}_{f\left({M}_{i}\right)}\left(y\right)=\sum _{{g}^{-1}\left(z\right)\ne \xd8g\left(y\right)=z}\sum _{{f}^{-1}\left(y\right)\ne \xd8f\left(x\right)=y}{C}_{{M}_{i}}\left(x\right)==\sum _{gf\left(x\right)=z}{C}_{{M}_{i}}\left(x\right)={C}_{gf\left({M}_{i}\right)}\left(z\right).$

**Example**

**6.**

## 5. Some Applications of Operations on Multiset

**Example**

**7.**

**Example**

**8.**

**Example**

**9.**

**Example**

**10.**

## 6. Conclusions

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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Hošková-Mayerová, Š.; Onasanya, B.O. Results on Functions on Dedekind Multisets. *Symmetry* **2019**, *11*, 1125.
https://doi.org/10.3390/sym11091125

**AMA Style**

Hošková-Mayerová Š, Onasanya BO. Results on Functions on Dedekind Multisets. *Symmetry*. 2019; 11(9):1125.
https://doi.org/10.3390/sym11091125

**Chicago/Turabian Style**

Hošková-Mayerová, Šárka, and Babatunde Oluwaseun Onasanya. 2019. "Results on Functions on Dedekind Multisets" *Symmetry* 11, no. 9: 1125.
https://doi.org/10.3390/sym11091125