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.

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- May 13th 2012, 05:30 AMsaravananbstopology
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. - May 13th 2012, 07:52 AMPlatoRe: topology
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? - May 13th 2012, 08:24 AMsaravananbsRe: topology
i can able to understand all your four statements.

after that i could't. - May 13th 2012, 08:30 AMPlatoRe: topology
- May 13th 2012, 08:34 AMsaravananbsRe: topology
if A is open in X then its complement is A' is finite. right!

for every open set O in Y there exist f^-1 (O) open in X => f is continuous. - May 13th 2012, 09:00 AMPlatoRe: topology