Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 24 / 6

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.