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Identify the space of infinite subsets of the natural numbers with their increasing enumerations, with the ordinary topology on infinite subsets inherited from the discrete product ${\omega }^{\omega }$. This is the sequence convention in the Open Ramsey theorem. For a general infinite ground set, first choose a countably infinite subset and use its enumeration. The theorem cannot mean unrestricted sequences with repetitions: the clopen partition according as the first two entries agree or differ has both colours over every infinite ground set.

For a finite finite stem $s$ and an infinite tail $B$ above $\max s$, writewhere the enumeration of $s$ precedes the tail. Let $O$ be open. The tail $B$ accepts $s$ if $[s,B]\subseteq O$; it rejects $s$ if no infinite subtail accepts it. Both properties persist on infinite subtails, and any stem can be decided by thinning: choose an accepting subtail if one exists; otherwise the current tail already rejects it.

Use deciding all finite stems by fusion. At stage $n$, choose a new element $b_n$ from the current tail, then thin the remaining tail to decide each of the finitely many subsets of $\{b_0,\dots ,b_n\}$. First decide the empty stem as well. The final selected set $B$ decides every finite stem from $B$, because all subsequent selections lie in the tail that decided that stem. If $B$ accepts the empty stem, $[B{]}^{\omega }\subseteq O$ and we are done.

Otherwise the empty stem is rejected. If a stem $s\subset B$ is rejected, only finitely many $a\in B$ above $\max s$ can have the tail $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed $D$, then every $X\in [D{]}^{\omega }$ starts with one of them and has its remaining tail in $B/a$. It follows that $[s,D]\subseteq O$, contradicting rejection of $s$. All other one-point extensions are rejected, since the first fusion has already decided them.

Build an infinite $C\subseteq B$ recursively. Maintain that every subset of the finite selected prefix is rejected. To select its next element, avoid the finite union of the finitely many accepting-extension obstructions for these rejected stems. Such an element exists, and all the new stems remain rejected. Hence every finite stem from $C$ is rejected. If some $X\in [C{]}^{\omega }$ were in $O$, openness would supply a finite initial segment $s$ of $X$ whose entire ordinary cylinder is contained in $O$. The tail $C/\max s$ would accept $s$, a contradiction. ThusIn particular every open payoff has an infinite homogeneous set. A finite partition into open colour classes can be handled by successive application of this two-colour assertion.

For an arbitrary partition use finite-symmetric-difference parity colouring. On the infinite subsets of the ground set, declare $X\sim Y$ when $X△Y$ is finite, and choose a representative $R$ from each equivalence class. Colour $X$ byFor any infinite $H$, removing one member gives an infinite $H^{\prime }$ in the same class and reverses the parity. Thus $H$ and $H^{\prime }$ have opposite colours, so neither colour contains all infinite subsets of any infinite ground set. This two-colouring has no infinite homogeneous set. Representative selection uses choice; the construction is deliberately unrestricted by topological regularity.

A BQO, or better-quasi-ordering, strengthens a well-quasi-ordering by testing arrays indexed by arbitrary barriers, rather than just sequences. Let $Q$ be a preorder. A barrier in better-quasi-order theory on an infinite $A\subseteq \omega$ is a family $B$ of nonempty finite increasing sequences such that no member is properly contained in another and every infinite subset of $A$ has a member of $B$ as its initial segment.

For barrier members write $s◃t$ if a finite increasing sequence $u$ has $s$ as an initial segment and has $t$ as an initial segment after deleting the first entry of $u$. A map $f:B\to Q$ is good when some $s◃t$ satisfy $f(s)\le f(t)$. The definition isBoth quantifiers matter: testing only barriers of one fixed finite length is insufficient.

For the singleton barrier $B=[A{]}^1$, shifted pairs are just $\{a\}◃\{b\}$ with $a<b$. Thus the BQO condition implies the ordinary well-quasi-ordering condition. Barrier arrays can also be viewed through continuous arrays in better-quasi-order theory $F:[A{]}^{\omega }\to Q$, with $Q$ discrete: set $F(X)=f(s)$ for the unique barrier initial segment $s$ of $X$. This is continuous because that value is fixed by a finite prefix, and comparing $X$ with $X\setminus \{\min X\}$ is exactly the shift comparison. The general continuous-array formulation is equivalent to the barrier definition; its converse uses refinement of fronts to barriers, rather than the incorrect assertion that every prefix-decision family is already a barrier.

For an illustrative positive example, every well-order is a BQO. If a barrier array into a well-order were bad, its induced $F$ would satisfyfor every infinite $X$. This is an infinite strict descent in a well-order, a contradiction. A finite antichain with equality is also a BQO: the finitely many fibres of the induced continuous array are open, so repeated use of the Open Ramsey theorem gives an infinite $A^{\prime }$ on which the array is constant. The comparison of an infinite subset of $A^{\prime }$ with its shift is then good.

The implication to a well-quasi-order is strict. The Rado order isIt is well-quasi-ordered: if one first coordinate occurs infinitely often, its second coordinates have an increasing pair; otherwise the first coordinates are unbounded, and a later one exceeds the second coordinate of a chosen earlier pair. But take the barrier $[\omega {]}^2$ and $f(\{m,n\})=(m,n)$. For $m<n<l$, the shift compares $(m,n)$ with $(n,l)$; their first coordinates differ and $n<n$ is false. Every shifted comparison therefore fails. This bad barrier array for the Rado order proves well-quasi-ordering does not imply better-quasi-ordering.

The stronger notion is designed for infinitary closure operations, such as powerset and infinite-sequence constructions, where a mere well-quasi-order can develop new bad arrays. The barrier definition records variable finite amounts of information from an infinite object and is consequently more robust than checking sequences alone.