Editing Ramsey's theorem (section)

==== Infinite version implies the finite ====
It is possible to deduce the finite Ramsey theorem from the infinite version by a [[proof by contradiction]]. Suppose the finite Ramsey theorem is false. Then there exist integers {{mvar|c}}, {{mvar|n}}, {{mvar|T}} such that for every integer {{mvar|k}}, there exists a {{mvar|c}}-colouring of {{math|[''k'']{{sup|(''n'')}}}} without a monochromatic set of size {{mvar|T}}. Let {{mvar|C{{sub|k}}}} denote the {{mvar|c}}-colourings of {{math|[''k'']{{sup|(''n'')}}}} without a monochromatic set of size {{mvar|T}}.

For any {{mvar|k}}, the restriction of a colouring in {{math|''C''{{sub|''k''+1}}}} to {{math|[''k'']{{sup|(''n'')}}}} (by ignoring the colour of all sets containing {{math|''k'' + 1}}) is a colouring in {{mvar|C{{sub|k}}}}. Define {{tmath|C^1_k}} to be the colourings in {{mvar|C{{sub|k}}}} which are restrictions of colourings in {{math|''C''{{sub|''k''+1}}}}. Since {{math|''C''{{sub|''k''+1}}}} is not empty, neither is {{tmath|C^1_k}}.

Similarly, the restriction of any colouring in {{tmath|C^1_{k+1} }} is in {{tmath|C^1_k}}, allowing one to define {{tmath|C^2_k}} as the set of all such restrictions, a non-empty set. Continuing so, define {{tmath|C^m_k}} for all integers {{mvar|m}}, {{mvar|k}}.

Now, for any integer {{mvar|k}},
:<math>C_k\supseteq C^1_k\supseteq C^2_k\supseteq \cdots</math>
and each set is non-empty. Furthermore, {{mvar|C{{sub|k}}}} is finite as
:<math>|C_k|\le c^{\frac{k!}{n!(k-n)!}}</math>
It follows that the intersection of all of these sets is non-empty, and let
:<math>D_k=C_k\cap C^1_k\cap C^2_k\cap \cdots</math>
Then every colouring in {{mvar|D{{sub|k}}}} is the restriction of a colouring in {{math|''D''{{sub|''k''+1}}}}. Therefore, by unrestricting a colouring in {{mvar|D{{sub|k}}}} to a colouring in {{math|''D''{{sub|''k''+1}}}}, and continuing doing so, one constructs a colouring of <math>\mathbb N^{(n)}</math> without any monochromatic set of size {{mvar|T}}. This contradicts the infinite Ramsey theorem.

If a suitable topological viewpoint is taken, this argument becomes a standard [[compactness theorem|compactness argument]] showing that the infinite version of the theorem implies the finite version.<ref>{{Cite book|title=Graph Theory|last=Diestel|first=Reinhard|publisher=Springer-Verlag|year=2010|isbn=978-3-662-53621-6|edition=4|location=Heidelberg|pages=209–2010|chapter=Chapter 8, Infinite Graphs}}</ref>