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Let $\sqcup$ denote the disjoint union. Then is there nice deMorgan relation we can obtain for:

$$\left(\bigsqcup_{i=1}^n E_i\right)^c$$

On the other hand, suppose we have the following

$$\bigcap_{i=1}^n \left(\bigsqcup_{j=1}^{m_i} B_{i_j}\right)$$

Is there a particularly pleasant distribution of unions and intersections that yields something like $\sqcup\cap$? So that ultimately I have a disjoint union (with respect to the outside set operation)?

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1 Answer 1

You can't define disjoint union inside a single set, but DeMorgan's law requires that. For example, if $S=\{1,2,3\}$ and $E_1=E_2=E_3=E_4=\{1\}$, what set are you using to perform the complement of $\bigsqcup E_i$? Is it different if you change $E_4=\emptyset$?

Disjoint unions are also problematic if you are dealing with intersections, since the set you get from a disjoint union is actually order-dependent - a different indexing of the sets gives a different order.

Basically, unlike unions, disjoint unions are really only unique up to isomorphism, and is not necessarily contained in any particular set.

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Well..... I guess the best we can do is the following (since all the intersections and unions are finite): $$\bigcap_{i=1}^n (\bigsqcup_{j=1}^{m_i} B_{i_j}) = \bigsqcup_{j=1}^k (\bigcap_{i=1}^n B_{i_j}) $$ where $k$ is some finite number and is given by the number of disjoint intersections. T. Andrews this seems sensible at least, right? –  Squirtle Sep 8 '13 at 2:21

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