 Solution

A well-quasi-ordering is a preorder $(Q,\le )$ such that every infinite sequence $(x_n)$ has indices $i<j$ with $x_i\le x_j$. Reflexivity and transitivity are part of the definition; antisymmetry is unnecessary. An infinite sequence without such a pair is a bad sequence.

The perfect subsequence lemma strengthens this definition: every infinite sequence in a well-quasi-order has an infinite nondecreasing subsequence, meaningHere is a proof that does not assume the lemma implicitly. There must be an index $i$ with infinitely many later indices $j$ satisfying $x_i\le x_j$. Otherwise each index would have only finitely many such successors. Starting with any index, choose the next one beyond the union of those finite successor sets for the previously chosen indices. This constructs a bad sequence, contradicting the well-quasi-ordering property.

Choose $i_0$ with infinitely many successors above it, and restrict to those successors. Apply the same argument to that infinite subsequence, obtaining $i_1>i_0$ with infinitely many later successors above it within the restricted sequence. Continue. Every subsequent restriction lies inside all previous upper cones, so $x_{i_r}\le x_{i_s}$ for every $r<s$. This proves the lemma, including for a genuine quasi-order with distinct equivalent elements. Both alternatives below use it.