Preimage of Intersection under Mapping/Family of Sets/Proof 2

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 mapping.

Then:

$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

where:

$f^{-1} \sqbrk {T_i}$ denotes the preimage of $T_i$ under $f$.

$\ds x$

$\in$

$\ds f^{-1} \sqbrk {\bigcap_{i \mathop \in I} T_i}$

$\ds \leadstoandfrom \ \ $

$\ds \map f x$

$\in$

$\ds \bigcap_{i \mathop \in I} T_i$

$\ds \leadstoandfrom \ \ $

$\ds \forall i \in I: \, $

$\ds \map f x$

$\in$

$\ds T_i$

$\ds \leadstoandfrom \ \ $

$\ds \forall i \in I: \, $

$\ds x$

$\in$

$\ds f^{-1} \sqbrk {T_i}$

$\ds \leadstoandfrom \ \ $

$\ds x$

$\in$

$\ds \bigcap_{i \mathop \in I} f^{-1} \sqbrk {T_i}$

$\blacksquare$

1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions: Theorem $7 \ \text{(e)}$
1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites