Does a circle have a subset which is topologically closed and which contains just one of each pair of diametrically opposite points?
The map taking each point to the diametrically opposite point is a homeomorphism. [It is obviously (1,1) and its own inverse. So it is sufficient to prove it continuous. But that is almost obvious using an ε δ argument.] Homeomorphisms take closed sets to closed sets. So if the subset was closed, then its complement would also be closed. So the circle would be the disjoint union of two closed sets and hence not connected. But it is connected. [The purist will rightly regard this statement as non-obvious, but it is pure bookwork.]
36th Putnam 1975
© John Scholes
27 Jan 2001