If is an infinite set with its cofinite topology and is a Hausdorff space, every continuous map is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of , contradicting its hyperconnected space property.
For example, from with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval is , which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.
Let be a nonempty infinite set with the cofinite topology, let be any topological space, and let be Hausdorff. For a continuous map , every slice is constant. Thus , where for any fixed . The map is continuous, since is continuous into the product topology. It is unique because the coordinate projection is a surjection. If is a surjection, so is .
For and homeomorphic to the circle, a surjective cannot be injective: otherwise the compact-to-Hausdorff continuous bijection theorem would contradict the fact that the circle and interval are not homeomorphic. Hence distinct satisfy simultaneously for all .
A continuous map from a nonempty finite product space of infinite sets with their cofinite topologies to a Hausdorff space is a constant function. Fixing all but one coordinate gives a continuous map from an infinite cofinite topology, so changing that coordinate does not change the value. Any two points in a finite Cartesian product can be joined by finitely many one-coordinate changes. This proves the assertion. In particular every real-valued continuous function on is constant when is the real line equipped with the cofinite topology.

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