Image of Doubleton under Mapping

Theorem

Let $S, T$ be sets.

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

Then:

$\forall x, y \in S: f \sqbrk {\set {x, y} } = \set {\map f x, \map f y}$

Proof

Let $x, y \in S$.

Thus

\(\ds f \sqbrk {\set {x, y} }\)

\(=\)

\(\ds f \sqbrk {\set x \cup \set y}\)

Union of Unordered Tuples

\(\ds \)

\(=\)

\(\ds f \sqbrk {\set x} \cup f \sqbrk {\set y}\)

Image of Union under Mapping

\(\ds \)

\(=\)

\(\ds \set {\map f x} \cup \set {\map f y}\)

Image of Singleton under Mapping

\(\ds \)

\(=\)

\(\ds \set {\map f x, \map f y}\)

Union of Unordered Tuples

$\blacksquare$

Sources

• Mizar article FUNCT_1:60