Union is Dominated by Disjoint Union

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Theorem

Let $I$ be an indexing set.

For all $i \in I$, let $S_i$ be a set.

Then:

$\displaystyle \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$

where $\preccurlyeq$ denotes domination, $\bigcup$ denotes union, and $\bigsqcup$ denotes disjoint union.


Proof

For all $\displaystyle x \in \bigcup_{i \mathop \in I} S_i$, there exists a $i \left({x}\right) \in I$ such that $x \in S_{i \left({x}\right)}$.

Thus the mapping $\displaystyle \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:

$\iota \left({x}\right) = \left({x, i \left({x}\right)}\right)$

is an injection.

$\blacksquare$