Preimage of Union under Mapping/Family of Sets

From ProofWiki
Jump to: navigation, search

Theorem

Let $S$ and $T$ be sets.

Let $\family {T_i}_{i \mathop \in I}$ be a family of subsets of $T$.

Let $f: S \to T$ be a relation.


Then:

$\displaystyle f^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

where:

$\displaystyle \bigcup_{i \mathop \in I} T_i$ denotes the union of $\family {T_i}_{i \mathop \in I}$
$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.


Proof 1

As $f$, being a mapping, is also a relation, we can apply Preimage of Union under Relation: Family of Sets:

$\displaystyle \mathcal R^{-1} \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \mathcal R^{-1} \sqbrk {T_i}$

where $\mathcal R^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $\mathcal R^{-1}$.

$\blacksquare$


Proof 2

We have that $f$ is a mapping, and so also a relation.

Thus its inverse $f^{-1}$ is also a relation.

Hence we can apply Image of Union under Relation: Family of Sets:

$\displaystyle \mathcal R \sqbrk {\bigcup_{i \mathop \in I} T_i} = \bigcup_{i \mathop \in I} \mathcal R \sqbrk {T_i}$

where $\mathcal R \sqbrk {T_i}$ denotes the image of $T_i$ under $\mathcal R$.

$\blacksquare$


Sources