Let "X" be a infinite set with cofinite topology. Let us prove that any bijection "f:X \\to X" is a homeomorphism. Let "U\\subset X" be an open set. By definition of cofinite topology, we have that "X\\setminus U" is finite. Since "f: X\\to X" is a bijection, we conclude that "f^{-1}(X\\setminus U)=f^{-1}(X)\\setminus f^{-1}(U)=X\\setminus f^{-1}(U)" and "f^{-1}( U)=X\\setminus f^{-1}(X\\setminus U)." Since "X\\setminus U" is finite, "f^{-1}(X\\setminus U)" is also finite, and hence "f^{-1}( U)" is cofinite. We conclude that "f^{-1}( U)" belongs to the cofinite topology, that is "f^{-1}( U)" is an open set. Therefore, "f" is a continuous map. By analogy, for any open set "V" its image "f(V)" is cofinite, and hence it is an open set in cofinite topology. We conclude that the map "f:X\\to X" is open, and hence "f" is a homeomorphism.
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