# Can a multiset be a subset of a set?

This is a bit of a silly question but its bothering me.

Given a multiset say $$p=\{a,a,g,h,h\}$$ And another set $$t=\{a,g,h\}$$ Can I say that $$p\subset t$$.

In other words is p a subset of t?.

• @KarthikKannan $p$ is a multiset. – Umberto P. Jan 22 at 11:51
• @karthik kannan yes but in the sense that p is just a collection of objects is it a subset or can we not relate multisets to sets? – The homeschooler Jan 22 at 11:52
• @UmbertoP. I apologize for not noticing that. I shall delete that comment. – Karthik Kannan Jan 22 at 11:53
• $p$ is not a submultiset of the (multi)set $t$ as some eleemnts occur too often. One might say that the underlying set of the multiset $p$ is a subset of the set $t$, perhaps? – Hagen von Eitzen Jan 22 at 12:09
• @Hagen von Eitzen is p not a subset of t then? – The homeschooler Jan 22 at 12:28

• A multiset is a pair $$(X, \sim)$$, with $$X$$ a set and $$\sim$$ an equivalence relation on $$X$$. In your case, one can make $$p=(P, \sim_{p})$$, with $$P=\{a, a^{*}, g, h, h^{*}\},$$ and $$a\sim_{p} a^{*}$$ and $$h\sim_{p} h^{*}$$ (meaning, in a way, that $$a$$ and $$a^{*}$$, and $$h$$ and $$h^{*}$$, are the same elements, $$a$$ and $$h$$ respectively). In that case, a set is a multiset where $$\sim$$ is the identity, and we say, for two multisets $$m_{1}=(X_{1}, \sim_{1})$$ and $$m_{2}=(x_{2}, \sim_{2})$$, that $$m_{1}$$ is contained in $$m_{2}$$ if $$X_{1}\subseteq X_{2}\quad\text{and}\quad \forall x_{1}, x_{2}\in X_{1} (x_{1}\sim_{1}x_{2}\Rightarrow x_{1}\sim_{2}x_{2})$$ (you could, instead of $$X_{1}\subseteq X_{2}$$, use that there is an injective function, but that would probably be considered more general). Then it is clear that, under this definition, $$p$$ is not a submultiset of $$t=(T, \sim_{t})$$, for $$T=\{a, g, h\}$$ and $$\sim_{t}$$ the identity on $$T$$; BUT, $$t$$ is a submultiset of $$p$$, and $$T$$ is a subset of $$P$$.
• A multiset is a pair $$(X, m)$$, with $$X$$ a set and $$m$$ a function mapping elements of $$X$$ to cardinals. That way, we can take $$p=(P, m_{p})$$, with $$P=\{a, g, h\}$$, and $$m_{p}(a)=2$$, $$m_{p}(g)=1$$ and $$m_{p}(h)=2$$ (meaning $$a$$ and $$h$$ appear twice in $$p$$, while $$g$$ appears once). We say $$(X_{1}, m_{1})$$ is a submultiset of $$(X_{2}, m_{2})$$ if $$X_{1}\subseteq X_{2}\quad\text{and}\quad \forall x\in X_{1}(m_{1}(x)\leq m_{2}(x)).$$ If we define $$t$$ as $$(T, m_{t})$$, for $$T=\{a, g, h\}$$ and $$m_{t}(a)=m_{t}(g)=m_{t}(h)=1$$ (a set is, with this definition, a multiset whose multiplicity function is always $$1$$), then we have $$t$$ is still a submultiset of $$p$$; but, and perhaps this definition is better regarding this intuitive aspect, $$P$$ and $$T$$ are EQUAL.
So, to summarize, you can not usually say $$p$$ is a submultiset of $$t$$; however, if you consider the underlying set of $$p$$ (the $$P$$ in my second definition), it indeed equals $$t$$.