Image under Inclusion Mapping
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Theorem
Let $X$ be a set.
Let $S \subseteq X$, $Z \subseteq S$.
Then $i_S \sqbrk Z = Z$
where
- $i_S$ denotes the inclusion mapping of $S$
- $i_S \sqbrk Z$ denotes the image of $Z$ under $i_S$.
Proof
\(\ds i_S \sqbrk Z\) | \(=\) | \(\ds \set {\map {i_S} z: z \in Z}\) | Definition of Image of Subset under Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \set {z: z \in Z}\) | Definition of Inclusion Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds Z\) | Definition of Set Equality |
$\blacksquare$
Sources
- Mizar article COMPACT1:2