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  1. MHB Master
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    #1
    Hey!!

    Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. Prove or disprove the following:
    1. $A$ closed $\Rightarrow$ $f(A)$ closed
    2. $B$ closed $\Rightarrow$ $f^{-1}(B)$ closed
    3. $A$ bounded $\Rightarrow$ $f(A)$ bounded
    4. $B$ bounded $\Rightarrow$ $f^{-1}(B)$ bounded



    I have done the following:

    We have that $D$ is closed $\iff$ for each convergent sequence $(x_n)$ in $D$ it holds that $\displaystyle{\lim_{n\rightarrow \infty} x_n\in D}$.

    We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

    1. What can we do in this case?
      $$$$
    2. Let $(x_n)$ be a convergent sequence in $f^{-1}(B)$. We want to show that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$.
      Since $f$ is continuous, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty} x_n\right )=\lim_{n\rightarrow \infty} f(x_n)}$.
      Since $(x_n)\in f^{-1}(B)$ and $f$ is continuous we have that $f(x_n)\in B$. Is this correct?
      Since $B$ is closed we have that $\displaystyle{\lim_{n\rightarrow \infty}f(x_n)\in B}$.
      Therefore, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty}x_n\right )\in B}$. From that it implies that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$. This means that $f^{-1}(B)$ is closed.
      Is this correct?
      $$$$
    3. Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$
      This means that in $A$ the function $f$ is bounded.
      Therefore, $f(A)$ is bounded.
      Is this correct?
      $$$$
    4. What can we do in this case?
    Last edited by mathmari; January 2nd, 2017 at 23:04.

  2. MHB Journeyman
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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. Prove or disprove the following:
    1. $A$ closed $\Rightarrow$ $f(A)$ closed
    2. $B$ closed $\Rightarrow$ $f^{-1}(B)$ closed
    3. $A$ bounded $\Rightarrow$ $f(A)$ bounded
    4. $B$ bounded $\Rightarrow$ $f^{-1}(B)$ bounded



    I have done the following:

    We have that $D$ is closed $\iff$ for each convergent sequence $(x_n)$ in $D$ it holds that $\displaystyle{\lim_{n\rightarrow \infty} x_n\in D}$.

    We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

    1. What can we do in this case?
      $$$$
    Consider $f(x)=\tan^{-1}(x)$. Then $f(\mathbf R)=(-\pi/2, \pi/2)$. Thus $(1)$ is not true.

    Quote Originally Posted by mathmari View Post

    1. Let $(x_n)$ be a convergent sequence in $f^{-1}(B)$. We want to show that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$.
      Since $f$ is continuous, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty} x_n\right )=\lim_{n\rightarrow \infty} f(x_n)}$.
      Since $(x_n)\in f^{-1}(B)$ and $f$ is continuous we have that $f(x_n)\in B$. Is this correct?
      Since $B$ is closed we have that $\displaystyle{\lim_{n\rightarrow \infty}f(x_n)\in B}$.
      Therefore, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty}x_n\right )\in B}$. From that it implies that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$. This means that $f^{-1}(B)$ is closed.
      Is this correct?
      $$$$
    This is ok. Thus (2) is true.

    Quote Originally Posted by mathmari View Post

    1. Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$ (no, this may not be true).
    2. This means that in $A$ the function $f$ is bounded.
      Therefore, $f(A)$ is bounded.
      Is this correct?
      $$$$
    One can show that the image of a closed interval under a continuous function $f:\mathbf R\to \mathbf R$ is again a alosed interval.
    From this it will follow that (3) is true.

    Quote Originally Posted by mathmari View Post

    1. What can we do in this case?
    As for (4), again take $f=\tan^{-1}$ and see that $f^{-1}(-\pi/2, \pi/2)=\mathbf R$. Thus (4) is false.

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    #3 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    One can show that the image of a closed interval under a continuous function $f:\mathbf R\to \mathbf R$ is again a alosed interval.
    Isn't this the case 1., that we disaproved?

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    #4 Thread Author
    Do you mean the following?

    Since $A$ is bounded, there exists $b=\sup A$ and $a=\inf A$. So, $A\subseteq [a,b]$.
    Since $f$ is continuous (or do we not need the continuity here ? ) we get that $f(A)\subseteq f([a,b])$.
    $[a,b]$ is closed and $f$ is continuous, therefore $f([a,b])$ is closed. How could we prove that?
    Is $f(A)$ then bounded because it is a subset of a closed interval?

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    #5
    Quote Originally Posted by mathmari View Post
    Do you mean the following?

    Since $A$ is bounded, there exists $b=\sup A$ and $a=\inf A$. So, $A\subseteq [a,b]$.
    Since $f$ is continuous (or do we not need the continuity here ? ) we get that $f(A)\subseteq f([a,b])$.
    $[a,b]$ is closed and $f$ is continuous, therefore $f([a,b])$ is closed. How could we prove that?
    Is $f(A)$ then bounded because it is a subset of a closed interval?
    In case I caused any confusion, by a closed interval I mean a subset of $\mathbf R$ of the form $[a, b]$, where $a<b$ are real numbers.

    The statement I made was that if $f:\mathbf R\to \mathbf R$ is a continuous function, then $f$ maps a closed interval to a closed interval.

    A one line proof of this would be to invoke the fact that any continuous image of a compact-connected topological space is again compact-connected.

    Here is a more basic proof using sequences.

    We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$. But this means $f(x_{n_k})$ is bounded which is a contradiction. Similarly one can prove that $B$ is bounded below.

    Let $x=\inf(B)$ and $y=\sup(B)$. Again, an application of Bolzano-Weierstrass shows that both $x$ and $y$ are in $B$. Now by intermediate value theorem it is clear that $B=[x, y]$.
    Last edited by caffeinemachine; January 3rd, 2017 at 21:23.

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    #6 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$, giving $N_k\to f(x)$. But this means $N_k$ is bounded which is a contradiction. Similarly one can prove that $B$ is bounded below.
    I haven't really understood the part with $N_k$. What exactly is $N_k$ ?

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    #7
    Quote Originally Posted by mathmari View Post
    I haven't really understood the part with $N_k$. What exactly is $N_k$ ?
    I have edited my post. (Actually, writing $N_k\to f(x)$ was not correct.)

    To elaborate a bit, a subsequence of $(x_n)_{n\geq 1}$ is a sequence $(x_{n_k})_{k\geq 1}$, where $(n_k)_{k\geq 1}$ is a strictly increasing sequence of naturals.

    Does that clear the confusion?

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    #8 Thread Author
    So, we use the fact that a convergent sequence is bounded, right?
    Having that $f(x_{n_k})$ is bounded, means that $f(x_n)$ must also be bounded?


    Quote Originally Posted by caffeinemachine View Post
    Let $x=\inf(B)$ and $y=\sup(B)$. Again, an application of Bolzano-Weierstrass shows that both $x$ and $y$ are in $B$. Now by intermediate value theorem it is clear that $B=[x, y]$.
    Let w be a number between x and y, then there must be at least one value c within [a, b] such that f(c) = w, right? How does it follow from that that $B=[x,y]$ ? I got stuck right now...


    In general, it holds that $B\subseteq [x,y]$, or not? Since $x,y\in B$, it follows that $B=[x,y]$, right?
    Last edited by mathmari; January 3rd, 2017 at 22:39.

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    #9
    Quote Originally Posted by mathmari View Post
    So, we use the fact that a convergent sequence is bounded, right?
    Yes.

    Quote Originally Posted by mathmari View Post
    Having that $f(x_{n_k})$ is bounded, means that $f(x_n)$ must also be bounded?
    No. You can come up with easy counterexamples to this.

    Quote Originally Posted by mathmari View Post
    Let w be a number between x and y, then there must be at least one value c within [a, b] such that f(c) = w, right?
    Yes.

    Quote Originally Posted by mathmari View Post
    How does it follow from that that $B=[x,y]$ ? I got stuck right now...

    In general, it holds that $B\subseteq [x,y]$, or not? Since $x,y\in B$, it follows that $B=[x,y]$, right?
    We defined $x=\inf(B)$ and $y=\sup(B)$. Thus we have $B\subseteq [x, y]$. The reverse inequality is by the intermediate value theorem.

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    #10 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    No. You can come up with easy counterexamples to this.
    Quote Originally Posted by caffeinemachine View Post
    We want to show that $B:=f([a, b])$ is a closed interval. First we show that $B$ is bounded above. Suppose not. Then for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$. Now by Bolzano-Weierstrass, the sequence $(x_n)$ has a convergent subsequence, say $(x_{n_k})$, which converges to say $x$. Thus $f(x_{n_k})\to f(x)$. But this means $f(x_{n_k})$ is bounded which is a contradiction.
    But how do we get the contradiction?

    Do we maybe have to suppose that for each sequence $x_n\in [a,b]$ there is a natural number $N$ such that $f(x_n)>N$ and not that for each natural number $N$, there is $x_n\in [a, b]$ such that $f(x_n)>N$?
    Last edited by mathmari; January 4th, 2017 at 09:00.

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