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Math Help - prove that T is closed.

  1. #1
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    prove that T is closed.

    Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
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    Quote Originally Posted by tn11631 View Post
    Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.

    Suppose x_0\in [0,1]\setminus T\Longrightarrow f(x_0)-g(x_0)=c\neq 0 \Longrightarrow since [/tex]f(x)=g(x)[/tex] is continuous on [0,1] we get that there exists \epsilon > 0 s.t.

    f(x)-g(x)\neq 0\,\,\,\forall\,x\in I_\epsilon:=(x_0-\epsilon,x_0+\epsilon)\Longrightarrow I_\epsilon\subset [0,1]\setminus T \Longrightarrow [0,1]\setminus T is open ...

    Tonio
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    Quote Originally Posted by tn11631 View Post
    Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
    Here is a second way.
    Suppose that  \left( {t_n } \right) is a sequence of points in T.
    If \left( {t_n } \right) \to z then showing that z\in T we have done the problem.
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    Quote Originally Posted by tonio View Post
    Suppose x_0\in [0,1]\setminus T\Longrightarrow f(x_0)-g(x_0)=c\neq 0 \Longrightarrow since [/tex]f(x)=g(x)[/tex] is continuous on [0,1] we get that there exists \epsilon > 0 s.t.

    f(x)-g(x)\neq 0\,\,\,\forall\,x\in I_\epsilon:=(x_0-\epsilon,x_0+\epsilon)\Longrightarrow I_\epsilon\subset [0,1]\setminus T \Longrightarrow [0,1]\setminus T is open ...

    Tonio
    Sorry I just want to clarify symbols. Where you have \ were using the "a-b" right? looking at it in this box i see that it says setminus which is what i thought. and then do we have to show that f(x)=g(x) is continuous on [0,1]?
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    I think the second way is a bit easier to do...

    Since both f and g are continuous, you know that for a sequence  \{ x_n \} such that x_n \rightarrow x_0 , we have  f(x_n) \rightarrow f(x_0) , g(x_n) \rightarrow g(x_0) .

    Now, think about this, if the sequence x_n is from T, then we should have  f(x_n) = g(x_n) , right? Well, you just need to prove that x_0 is also in T. In other words, you need something like f(x_0)=g(x_0), which is true, but why?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
    More generally the follow works whenever the codomain is Hausdorff.


    Consider that A(f,g)=\left\{x\in[a,b]:f(x)=g(x)\right\}=\varphi^{-1}(\Delta) where \Delta is the diagonal and \varphi:[a,b]\mapsto\mathbb{R}\times\mathbb{R} given by x\mapsto(f(x),g(x))
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  7. #7
    Senior Member Tinyboss's Avatar
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    T is the inverse image of {0} (a closed set in R) under f-g (a continuous map).
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    MHF Contributor Drexel28's Avatar
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    As a note of interest this shows that if A(f,g) contains a dense subset of \text{Dom }f then f=g
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