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Math Help - real hard problem

  1. #1
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    real hard problem

    Consider a function f:\mathbb{R}\rightarrow\mathbb{R}. Prove f is continuous on \mathbb{R} iff for every sunbset A of \mathbb{R} f(\overline{A})\subseteq\overline{f(A)}
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  2. #2
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    Quote Originally Posted by mancillaj3 View Post
    Consider a function f:\mathbb{R}\rightarrow\mathbb{R}. Prove f is continuous on \mathbb{R} iff for every sunbset A of \mathbb{R} f(\overline{A})\subseteq\overline{f(A)}
    Lemma 1. f is continuous iff for each closed subset C of \mathbb{R} (codomain of f), f^{-1}(C) is closed in \mathbb{R} (domain of f).

    ->
    Assume f is continuouus and let A be a subset of \mathbb{R} (domain of f). Then, \overline{f(A)} is a closed subset of \mathbb{R} (codomain of f), so its inverse f^{-1}(\overline{f(A)}) is closed in \mathbb{R} (domain of f) by lemma 1.
    Since A \subset f^{-1}(\overline{f(A)}) (the latter set is closed), we have \overline{A} \subset f^{-1}(\overline{f(A)}).
    Thus, f(\overline{A})\subseteq\overline{f(A)}.

    <-
    Assume f(\overline{A})\subseteq\overline{f(A)} is true for each subset A of \mathbb{R} (domain of f).
    Let C be a closed subset of \mathbb{R} (codomain of f). Then, f(\overline{f^{-1}(C)}) \subset \overline{ff^{-1}(C)} \subset \overline{C}=C. Thus, \overline{f^{-1}(C)} \subset f^{-1}(C), which implies that f^{-1}(C) is a closed subset in \mathbb{R} (domain of f).
    We conclude that f is continuous by lemma 1.
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  3. #3
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    Quote Originally Posted by mancillaj3 View Post
    Consider a function f:\mathbb{R}\rightarrow\mathbb{R}. Prove f is continuous on \mathbb{R} iff for every sunbset A of \mathbb{R} f(\overline{A})\subseteq\overline{f(A)}
    Here is another proof:

    For that among other theorems we will use the following theorem:

    Given any subset A of the real Nos R AND xεR,THEN:

    x\in\overline A iff there exists a sequence \{x_{n}\} in A and such that \lim_{n\rightarrow\infty}{x_{n}}= x

    Now let y\in f(\overline A)\Longrightarrow y=f(x)\wedge x\in\overline A=====> THERE exists a sequence \{x_{n}\} in A and \lim_{n\rightarrow\infty}{x_{n}}=x

    BUT f is also continuous over R AND hence over A AND thus \lim_{n\rightarrow\infty}{f(x_{n})} = f(x)

    Also \{x_{n}\} is in A HENCE f(x_{n}) is in f(A) AND so f(x)\in\overline {f(A)}

    Thus f(\overline A)\subset\overline {f(A)}

    Conversely:

    Suppose f(\overline A)\subset\overline {f(A)} and also that f is not continuous over R.

    Then there exist xεR AND a sequence in R \{x_{n}\} such that \lim_{n\rightarrow\infty}{x_{n}}=x and \lim_{n\rightarrow\infty}{f(x_{n})}\neq f(x).................................................. ............................1.

    Now put  A= \{x_{n}: n\in N\}=\{x_{n}\},hence x\in\overline A\Longrightarrow f(x)\in\ f(\overline A)\Longrightarrow f(x)\in\overline{f(A)}.

    BUT by the construction of A WE have:

    \lim_{n\rightarrow\infty}{f(x_{n})}=f(x).................................................. ...........................2

    (1) and (2) lead us to a contradiction,hence f is continuous over R
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  4. #4
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    I tried to give another proof by using the ε-δ definitions only,so

    Let w\in f(\overline A)\Longrightarrow w=f(x) ,x\in\overline A,hence x belongs to the real Nos R

    Let ε>0

    Since f is continuous over R AND thus at xεR we have :

    There exists δ>0 and such that:

    \forall z,  |z-x|<\delta\Longrightarrow |f(x)-f(z)|<\epsilon.................................................. ........................1

    Since x\in\overline A ,for any +ve No and thus for δ>0,there exists a ,y and such that:


    yεB(x,δ) and yεA,or

    |x-y|<δ and yεΑ............................................... ............................2

    put now z=y in (1) and

     |x-y|<\delta\Longrightarrow |f(x)-f(y)|<\epsilon and since yεA =====> f(y)εf(A) WE have ,by using (2):


    |f(x)-f(y)|<ε and f(y)εf(A) ,SO

    FOR all ε>0 ,then f(y)εB(f(x),ε) and f(y)εf(A).HENCE

    w=f(x)\in\overline{f(A)}.

    For the converse i could not find a proof without using sequences or closed sets
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