if (R,T1)=X, (R,T2)=Y
where T1= co-finite topology
and T2= usual topology
if there exist a function f :X--> Y , which is continous, then prove that f is constant function.
please explain.
Suppose that $\displaystyle p\ne q,~\&~\{p,q\}\subset f(X).$ i.e the image has more than one point.
In the usual topology there are two disjoint open sets $\displaystyle p\in O~\&~q\in Q~.$
Because $\displaystyle f^{-1}(O)$ is open in the co-finite topology then its complement is finite.
That means that $\displaystyle f$ maps all but a finite collection of points into $\displaystyle O$.
What is wrong with that picture?
If that is a contradiction, what does it prove?