# Thread: topology

1. ## topology

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.

2. ## Re: topology

Originally Posted by saravananbs
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.
Suppose that $p\ne q,~\&~\{p,q\}\subset f(X).$ i.e the image has more than one point.

In the usual topology there are two disjoint open sets $p\in O~\&~q\in Q~.$

Because $f^{-1}(O)$ is open in the co-finite topology then its complement is finite.

That means that $f$ maps all but a finite collection of points into $O$.

What is wrong with that picture?

If that is a contradiction, what does it prove?

3. ## Re: topology

i can able to understand all your four statements.
after that i could't.

4. ## Re: topology

Originally Posted by saravananbs
i can able to understand all your four statements. after that i could't.
That is an incoherent statement.

Tell us, what is the definition of cofinite topology?

What it mean to say that a function is continuous in terms of a topology?

5. ## Re: 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.

6. ## Re: topology

Originally Posted by saravananbs
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.
Those are correct.
Let $U=\left[f^{-1}(O)\right]^/~\&~V=\left[f^{-1}(Q)\right]^/$.
Those are both finite sets. Correct?
Now $f(X\setminus U)\subset O~\&~f(X\setminus V)\subset Q$. Is that possible?