Proof
(related to Proposition: Well-ordered Sets are Chains)
- Suppose $(V,\preceq )$ is well-ordered.
- We have to prove that "$\preceq$" is a total order, i.e. reflexive, transitive, antisymmetric, and connex.
- Since "$\preceq$" is a partial order, and since any partial order is reflexive, transitive and antisymmetric, it only remains to be shown that "$\preceq$" is connex.
- Since $V$ is well-ordered, any of its non-empty, finite subsets $S$ has a minimum.
- In particular, if $S=\{x,y\}$ consists of exactly two arbitrary elements of $V$, it has a minimum, and therefore we have $x\preceq y$ or $y\preceq x$.
- Since $x,y$ are arbitrarily chosen from $V$, "$\preceq$" is connex, as required.
- Altogether, "$\preceq$" is a total order and $V$ a chain, by definition.
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References
Bibliography
- Reinhardt F., Soeder H.: "dtv-Atlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition
