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 such that for every integer, there exists a -colouring of without a monochromatic set of size . Let denote the -colourings of without a monochromatic set of size .
For any k, the restriction of a colouring in to (by ignoring the colour of all sets containing ) is a colouring in . Define to be the colourings in which are restrictions of colourings in . Since is not empty, neither is .
Similarly, the restriction of any colouring in is in, allowing one to define as the set of all such restrictions, a non-empty set. Continuing so, define for all integers .
Now, for any integer, and each set is non-empty. Furthermore, is finite as . It follows that the intersection of all of these sets is non-empty, and let . Then every colouring in is the restriction of a colouring in . Therefore, by unrestricting a colouring in to a colouring in, and continuing doing so, one constructs a colouring of without any monochromatic set of size . This contradicts the infinite Ramsey theorem.
If a suitable topological viewpoint is taken, this argument becomes a standard compactness argument showing that the infinite version of the theorem implies the finite version.
Read more about this topic: Ramsey's Theorem
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