# Proof

(related to Proposition: Subsets of Finite Sets)

• If $$S\subseteq X$$ and $$S=X$$ then $$S$$ is finite, since $$X$$ is finite by hypothesis.
• Let $$S\subset X$$ and $$S\neq X$$.
• Since $$X$$ is finite by hypothesis, there is an $$n\in\mathbb N$$ with $$|X|=|n|$$.
• Since $$S\subset X$$, $$|S| < |X|$$, by characterization of finite sets $(3)$ and comparison of cardinal numbers.
• Let $$K$$ be the set of all natural numbers, with $$|S| \le k, k\in K$$. Since $$n\in K$$, the set $$K$$ is non-empty.
• Following the well-ordering principle, $$K$$ has a minimal element $$k_0\in K$$, for which $$|S| \le k_0$$.
• Note, that $$|S| \not < k_0$$. Otherwise for some $$j < k_0, j\in K$$ we would have $$|S| \le j$$, contradicting the minimality of $$k_0$$.
• Therefore $$|S| = k_0=|k_0|$$.
• Thus, $$S$$ is finite.

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### References

#### Bibliography

1. Reinhardt F., Soeder H.: "dtv-Atlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition