Set Finite iff Surjection from Initial Segment of Natural Numbers
Theorem
Let $S$ be a set.
Then $S$ is finite if and only if for some $n \in \N$ there exists a surjection $f: \N_{< n} \to S$.
Here, $\N_{< n}$ denotes an initial segment of $\N$.
Proof
Necessary Condition
Suppose that $S$ is finite.
By definition, this means there exists a bijection $f: \N_{< n} \to S$.
Then $f$ is a fortiori also the sought surjection.
$\Box$
Sufficient Condition
Let $f: \N_{< n} \to S$ be a surjection.
Define $g: S \to \N_{< n}$ by:
- $g (s) := \min f^{-1} (s)$
where $f^{-1} (s)$ is the preimage of $s$ under $f$.
Note that $f^{-1} (s)$ is not empty because $f$ is a surjection.
By the Well-Ordering Principle, $f^{-1} (s) \subseteq \N$ has a smallest element.
Hence $g$ is well-defined.
Next we show that $g$ is injective.
So suppose that $g(s) = g(s')$ for some $s, s' \in S$:
\(\ds g(s)\) | \(=\) | \(\ds g(s')\) | ||||||||||||
\(\ds \implies \ \ \) | \(\ds f \left({ g(s) }\right)\) | \(=\) | \(\ds f \left({ g(s') }\right)\) | |||||||||||
\(\ds \implies \ \ \) | \(\ds s\) | \(=\) | \(\ds s'\) | Definition of $g$ |
Hence $g$ is injective.
Then by Injection to Image is Bijection, $S$ is equivalent to a subset of $\N_{<n}$.
By Subset of Finite Set is Finite, it follows that $S$ is finite.
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.7$