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  1. MHB Master
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    #11 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    Consider $f(x)=\tan^{-1}(x)$. Then $f(\mathbf R)=(-\pi/2, \pi/2)$. Thus $(1)$ is not true.
    How can we show that $f(\mathbf R)=(-\pi/2, \pi/2)$ ?

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    #12 Thread Author
    Quote Originally Posted by mathmari View Post
    3. $A$ bounded $\Rightarrow$ $f(A)$ bounded
    Could we maybe show it also as follows?

    We suppose that $f(A)$ is not bounded.

    There is also a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$. Or do we not get that when $f(A)$ is not bounded?

    Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ it follows that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$, or not?

    $\{f^{-1}(x_n)\}_{n=1}^{\infty}$ is therefore, bounded, right?

    From Bolzano-Weierstrass there is a convergent subsequence, let $\{f^{-1}(x_{n_k})\}_{k=1}^{\infty}$ with $f^{-1}(x_{n_k})\rightarrow y\in \mathbb{R}$.

    Since $f$ is continuous, we have that $f\left (f^{-1}(x_{n_k})\right )\rightarrow f(y)\in \mathbb{R}$.

    That means that $x_{n_k}\rightarrow f(y)$.

    But how could we get a contradiction?
    Last edited by mathmari; January 5th, 2017 at 05:13.

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    #13
    Quote Originally Posted by mathmari View Post
    Could we maybe show it also as follows?

    We suppose that $f(A)$ is not bounded.

    There is also a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$. Or do we not get that when $f(A)$ is not bounded?
    It could be that there is no sequence in $f(A)$ whose limit is infinity, given that $f(A)$ is unbounded. But what is true for sure is that that is a sequence whose limit is either $+\infty$ or $-\infty$. We may WLOG assume that the limit is $+\infty$.

    Quote Originally Posted by mathmari View Post
    Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ it follows that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$, or not?
    Note that $f^{-1}(x_n)$ is a subset of of $\mathbf R$. So $\{f^{-1}(x_n)\}$ is not a sequence in $\mathbf R$.
    So the way to go is by choose $y_n\in f^{-1}(x_n)$ for each $n$. Then you get a sequence $\{y_n\}$ in $\mathbf R$.

    Now from Bolzano-Weierstrass there is a convergent subsequence $\{y_{n_k}\}_{k=1}^{\infty}$ with $y_{n_k}\rightarrow y\in \mathbb{R}$.

    Since $f$ is continuous, we have that $x_n=f(y_n)\rightarrow f(y)\in \mathbb{R}$.

    That means that $x_{n_k}\rightarrow f(y)$. The contradiction is that since $x_{n_k}$ is a subsequence of $\{x_n\}$, we also have $x_{n_k}\to \infty$.

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    #14 Thread Author
    I see!! Thank you very much!!

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