Editing Ramsey's theorem (section)

=== Case of more colours ===
''Lemma 2.'' If {{math|''c'' > 2}}, then <math>R(n_1, \dots, n_c) \leq R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c)).</math>

''Proof.'' Consider a complete graph of <math>R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c))</math> vertices and colour its edges with {{mvar|c}} colours. Now 'go colour-blind' and pretend that {{math|''c'' − 1}} and {{mvar|c}} are the same colour. Thus the graph is now {{math|(''c'' − 1)}}-coloured. Due to the definition of <math>R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c)),</math> such a graph contains either a {{mvar|K{{sub|n{{sub|i}}}}}} mono-chromatically coloured with colour {{mvar|i}} for some {{math|1 ≤ ''i'' ≤ ''c'' − 2}} or a {{math|''K''{{sub|''R''(''n''{{sub|''c'' − 1}}, ''n{{sub|c}}'')}}}}-coloured in the 'blurred colour'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of {{math|''R''(''n''{{sub|''c'' − 1}}, ''n{{sub|c}}'')}} we must have either a {{math|(''c'' − 1)}}-monochrome {{math|''K''{{sub|''n''{{sub|''c'' − 1}}}}}} or a {{mvar|c}}-monochrome {{mvar|K{{sub|n{{sub|c}}}}}}. In either case the proof is complete.

Lemma 1 implies that any {{math|''R''(''r'',''s'')}} is finite. The right hand side of the inequality in Lemma 2 expresses a Ramsey number for  {{mvar|c}} colours  in terms of Ramsey numbers for fewer colours. Therefore, any {{math|''R''(''n''{{sub|1}}, …, ''n{{sub|c}}'')}} is finite for any number of colours. This proves the theorem.