Properties of multiset union

June 4, 2008

We defined the union of a family of multisets last week. Today we’re going to look at some basic properties of the union operation on \mathbf{MSet}_X.

Suppose \left\{\mathcal{M}_i = \left(X, f_i\right)\right\}_{i\in I} is a family of multisets over X. Then the following relationships hold:

(i) \mathrm{support}\left(\bigcup_{i\in I}{\mathcal{M}_i}\right) = \bigcup_{i\in I}{\mathrm{support}\left(\mathcal{M}_i\right)}.

(ii) \mathcal{M}_i \subseteq \bigcup_{i\in I}{\mathcal{M}_i} for all i\in I.

(iii) \bigcup_{i\in I}{\mathcal{M}_i} \subseteq \biguplus_{i\in I}{\mathcal{M}_i}.

(iv) \bigcup_{i\in I}{\mathcal{M}_i} \subseteq \:\cdot\!\!\!\!\!\!\;\bigcup_{i\in I}{\mathcal{M}_i} provided that all of the multisets \mathcal{M}_i have the same support.

(v) \bigcap_{i\in I}{\mathcal{M}_i} \subseteq \bigcup_{i\in I}{\mathcal{M}_i}.

The proof of part (i), like the similar result for multiset intersections, is a straightforward series of equivalencies:

\begin{array}{r@{\:\Leftrightarrow\:}l} x\in \mathrm{support}\left(\bigcup_{i\in I}{\mathcal{M}_i}\right) & \sup f_i(x) \neq 0 \\ & \left(\exists i\in I\right)\: f_i\left(x\right)\neq 0 \\ & \left(\exists i\in I\right)\: x\in\mathrm{support}\left(\mathcal{M}_i\right) \\ & x\in\bigcup_{i\in I}{\mathrm{support}\left(\mathcal{M}_i\right)}. \end{array}

Part (ii) follows directly from the definition of the supremum of a set, since f_i\left(x\right) \leq \sup f_i\left(x\right) for all x\in X and i\in I.

To prove part (iii), start by recalling that the sum of a family of cardinals \left\{\kappa_i\right\}_{i\in I} is given by

\sum_{i\in I}{\kappa_i} = \left|\bigcup_{i\in I}{A_i}\right|

where the sets A_i are disjoint and \kappa_i = \left|A_i\right| for all i\in I. Therefore, letting A_{i,x} be a family of disjoint sets such that \left|A_{i,x}\right| = f_i\left(x\right), we have

\kappa_x = \sum_{i\in I}{f_i\left(x\right)} = \left|\bigcup_{i\in I}{A_{i,x}}\right| \geq \left|A_{i,x}\right| = f_i\left(x\right)

for all i\in I and x\in X. This establishes that \kappa_x is an upper bound for f_i\left(x\right) in the linearly ordered set \left(\mathbf{Card}, \leq\right), and so \kappa_x \geq \sup f_i\left(x\right). The result in part (iii) follows immediately since this holds for all x\in X.

The proof of part (iv) is similar, but relies on the definition of the product of a family of cardinals. The product of a family of cardinals \left\{\kappa_i\right\}_{i\in I} is given by

\prod_{i\in I}{\kappa_i} = \left|\prod_{i\in I}{A_i}\right|

where \kappa_i = \left|A_i\right| for all i\in I. Letting A_{i,x} be a family of sets such that \left|A_{i,x}\right| = f_i\left(x\right), we have

\kappa_x = \prod_{i\in I}{f_i\left(x\right)} = \left|\prod_{i\in I}{A_{i,x}}\right| \geq \left|A_{i,x}\right| = f_i\left(x\right)

for all i\in I and x\in X, provided that A_{i,x} \neq \emptyset for all i\in I. This establishes that \kappa_x is an upper bound for f_i\left(x\right) in the linearly ordered set \left(\mathbf{Card}, \leq\right) whenever x\in\mathrm{support}\left(\bigcup_{i\in I}{\mathcal{M}_i}\right), which is the common support of all multisets in the family \left\{\mathrm{M}_i\right\}_{i\in I}. If x is not in this support, then \prod_{i\in I}{f_i\left(x\right)} = 0 = \sup f_i\left(x\right). In either case, \kappa_x \geq \sup f_i\left(x\right), and part (iv) follows immediately since this holds for all x\in X.

The requirement that all multisets in the family have the same support is necessary; you can see this by considering the product and union of the multisets \left\{a\right\} and \left\{b\right\}.

Finally, part (v) follows immediately from the fact that

\inf f_i\left(x\right) \leq f_i\left(x\right) \leq \sup f_i\left(x\right)

for all i\in I and x\in X.

Copyright © 2008 Michael L. McCliment.

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