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Math Help - topology

  1. #1
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    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.
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  2. #2
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    Re: topology

    Quote Originally Posted by saravananbs View Post
    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?
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  3. #3
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    Re: topology

    i can able to understand all your four statements.
    after that i could't.
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  4. #4
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    Re: topology

    Quote Originally Posted by saravananbs View Post
    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?
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  5. #5
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    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.
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  6. #6
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    Re: topology

    Quote Originally Posted by saravananbs View Post
    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?
    Thanks from saravananbs
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