u is the usual topology, cf is the cofinite topology. yes proof: pick a and b in (R,cf) ((0,1),u) ~ ((a,b),u). then the identity on (a,b) is continuous is because (R,cf) [itex] \subset [/itex] (R,u). map 0->a and 1->b. the fn is continuous at the end points because no subset of the image is open in (R,cf). therefore their exists a path between any a and b in (R,cf). is this a valid proof?