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Thread: real hard problem

  1. #1
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    real hard problem

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

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

    <-
    Assume $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$ is true for each subset A of $\displaystyle \mathbb{R}$ (domain of f).
    Let C be a closed subset of $\displaystyle \mathbb{R}$ (codomain of f). Then, $\displaystyle f(\overline{f^{-1}(C)}) \subset \overline{ff^{-1}(C)} \subset \overline{C}=C$. Thus, $\displaystyle \overline{f^{-1}(C)} \subset f^{-1}(C)$, which implies that $\displaystyle f^{-1}(C)$ is a closed subset in $\displaystyle \mathbb{R}$ (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 $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$. Prove $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$ iff for every sunbset A of $\displaystyle \mathbb{R}$ $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$
    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:

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

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

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

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

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

    Conversely:

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

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

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

    BUT by the construction of A WE have:

    $\displaystyle \lim_{n\rightarrow\infty}{f(x_{n})}=f(x)$.................................................. ...........................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 $\displaystyle w\in f(\overline A)\Longrightarrow w=f(x) ,x\in\overline A$,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:

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

    Since $\displaystyle x\in\overline A$ ,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

    $\displaystyle |x-y|<\delta\Longrightarrow |f(x)-f(y)|<\epsilon$ 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

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

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