Successor Mapping on Natural Numbers is not Surjection
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Theorem
Let $f: \N \to \N$ be the successor mapping on the natural numbers $\N$:
- $\forall n \in \N: \map f n = n + 1$
Then $f$ is not a surjection.
Proof
There exists no $n \in \N$ such that $n + 1 = 0$.
Thus $\map f 0$ has no preimage.
The result follows by definition of surjection.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.2$