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  1. #1
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    help me in Prove

    Prove

    1) $\displaystyle X $ is a regular space, $\displaystyle A $ compact set and $\displaystyle B$ closed set,
    $\displaystyle A \bigcap B = {\O} $
    Then there is open sets U and V exist, such that
    $\displaystyle A\subseteq U, \ B \subseteq V,\ A \bigcap B = {\O} $


    2) $\displaystyle X $ is compact and hausdorff space and $\displaystyle f : X \rightarrow X $ is continuous function
    prove there is a closed set non empty such that $\displaystyle f(A)= A$
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  2. #2
    Senior Member roninpro's Avatar
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    For part two, define a sequence of sets in the following way: $\displaystyle A_0=X, A_1=f(A_0), A_2=f(A_1), A_3=f(A_2),\ldots$. Since $\displaystyle f:X\to X$ is continuous and takes a compact space to a Hausdorff space, the closed map lemma applies; we can say that each $\displaystyle A_i$ is a closed set. Let $\displaystyle A=\bigcap A_i$. Note that $\displaystyle A$ is closed and $\displaystyle f(A)=A$. To complete the proof, it necessary to show that $\displaystyle A$ is nonempty. I can't see how to do that at the moment, so hopefully somebody else can fill in the gap.

    I hope that this gives you some ideas, at any rate.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nice rose View Post
    Prove

    1) $\displaystyle X $ is a regular space, $\displaystyle A $ compact set and $\displaystyle B$ closed set,
    $\displaystyle A \bigcap B = {\O} $
    Then there is open sets U and V exist, such that
    $\displaystyle A\subseteq U, \ B \subseteq V,\ A \bigcap B = {\O} $
    Oh come on!

    Hint:

    Spoiler:
    For each $\displaystyle a\in A$ there are disjoint neighborhood $\displaystyle U_a,V_a$ with $\displaystyle a\in U_a$ and $\displaystyle B\subseteq V_a$. Cover $\displaystyle A$ with the set of all the $\displaystyle U_a$'s and procure a finite subcover. What then?


    2) $\displaystyle X $ is compact and hausdorff space and $\displaystyle f : X \rightarrow X $ is continuous function
    prove there is a closed set non empty such that $\displaystyle f(A)= A$
    Let's see some work

    Hint:

    Spoiler:


    What happens if $\displaystyle K=\left\{x\in X:f(x)\ne x\right\}=X$?
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