Proof
(related to Proposition: Wellordered Sets are Chains)
 Suppose $(V,\preceq )$ is wellordered.
 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 wellordered, any of its nonempty, 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.: "dtvAtlas zur Mathematik", Deutsche Taschenbuch Verlag, 1994, 10th Edition