Image of Subset under Mapping/Examples/Arbitrary Finite Set
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Example of Images of Subsets under Mapping
Let:
| \(\ds S\) | \(=\) | \(\ds \set {0, 1, 2, 3, 4, 5}\) | ||||||||||||
| \(\ds T\) | \(=\) | \(\ds \set {0, 1, 2, 3}\) |
Let $f: S \to S$ be the mapping defined as:
| \(\ds f \paren 0\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds f \paren 1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds f \paren 2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds f \paren 3\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds f \paren 4\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds f \paren 5\) | \(=\) | \(\ds 3\) |
Let:
| \(\ds A\) | \(=\) | \(\ds \set {0, 3}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \set {0, 1, 3}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \set {0, 1, 2}\) |
Then:
| \(\ds f \sqbrk A\) | \(=\) | \(\ds \set {0, 1}\) | ||||||||||||
| \(\ds f \sqbrk B\) | \(=\) | \(\ds \set {0, 1}\) | ||||||||||||
| \(\ds f \sqbrk C\) | \(=\) | \(\ds \set 0\) |
and:
- $\Img f = \set {0, 1, 3}$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 21.1$: The image of a subset of the domain; surjections