Results 1 to 8 of 8
Like Tree4Thanks
  • 1 Post By Idea
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By Idea

Thread: Mapping

  1. #1
    Junior Member
    Joined
    Aug 2018
    From
    Denmark
    Posts
    27

    Mapping

    Hey,

    Given that $\displaystyle f: X \rightarrow Y$ and that $\displaystyle A\subset X$ and $\displaystyle B \subset Y$ then I have to show that:

    $\displaystyle f(f^{-1}(B)) \varsubsetneq B $.

    I have done the following:

    Assume that $\displaystyle y \in f(f^{-1}(B)) \Leftrightarrow$
    $\displaystyle x \in f^{-1}(B)$ such that $\displaystyle f(x)=y \Leftrightarrow$
    $\displaystyle y=f(x) \in B \Leftrightarrow$
    $\displaystyle f(f^{-1}(B)) \subset B$

    What remains is to show that $\displaystyle f(f^{-1}(B)) \neq B $. Would a counterexample be sufficient?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,017
    Thanks
    519

    Re: Mapping

    Show that

    if $\displaystyle B\subseteq f\left(X\right)$ then $\displaystyle f\left(f^{-1}(B)\right)=B$
    Last edited by Idea; Jul 12th 2019 at 12:39 PM.
    Thanks from detalosi
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,405
    Thanks
    3296
    Awards
    1

    Re: Mapping

    Quote Originally Posted by detalosi View Post
    Given that $\displaystyle f: X \rightarrow Y$ and that $\displaystyle A\subset X$ and $\displaystyle B \subset Y$ then I have to show that:
    $\displaystyle f(f^{-1}(B)) \varsubsetneq B $.
    Suppose $X=\{0,2,4,6,8\}~\&~Y=\{0,1,\cdots 9\}$ if $f(x)=x+1~\&~B=\{1,3,5\}$
    $f\left(f^{-1}(B)\right)=?$
    Thanks from detalosi
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2018
    From
    Denmark
    Posts
    27

    Re: Mapping

    Yes I see that.
    So I have to show that

    $\displaystyle B \subseteq f(f^{-1}(B)) $

    does not hold.

    I have followed the same "method" as before by assuming that $\displaystyle y \in B$ (and then concluding that $\displaystyle y \notin f(f^{-1}(B))$ but it does not lead towards what is desired!

    How to move forward?

    Plato, yes I understand the example. But I wish to move forward without a counterexample.
    Last edited by detalosi; Jul 12th 2019 at 01:31 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,405
    Thanks
    3296
    Awards
    1

    Re: Mapping

    Quote Originally Posted by detalosi View Post
    Yes I see that.
    So I have to show that

    $\displaystyle B \subseteq f(f^{-1}(B)) $ does not hold.
    NOTATION: In place of $f(f^{-1}(B))$ I use $\overrightarrow f (\overleftarrow f (B))$
    Using the function I posted before, If $C=\{7,8,9\}$ then $\overleftarrow f (C)=\{6,8\}$
    while $\overrightarrow f (\overleftarrow f (C))=\{7,9\}\ne C$
    Are you sure you understand what exactly you are to show?
    Thanks from detalosi
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2018
    From
    Denmark
    Posts
    27

    Re: Mapping

    If $\displaystyle f(x)=x+1$ then $\displaystyle f^{-1}(x)=x-1$.

    Then $\displaystyle f^{-1}(B) = \{ 0,2,4 \}$.

    This means that $\displaystyle f(f^{-1}(B))=f(\{ 0,2,4 \})= \{ 1,3,5 \}$

    Notation might be off.
    Yes, I understand what I am trying to show. My thought process is as follows:

    1) I want to show that $\displaystyle f(f^{-1}(B) \varsubsetneq B$

    2) I have shown that $\displaystyle f(f^{-1}(B) \subseteq B$

    With my question I simply intended to show that = does not hold and therefore the inclusion is strict??
    Last edited by detalosi; Jul 12th 2019 at 02:34 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Aug 2018
    From
    Denmark
    Posts
    27

    Re: Mapping

    Have I done enough to show what was intended?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,017
    Thanks
    519

    Re: Mapping

    you will not be able to show that

    $$y\in B \Longrightarrow y\notin f\left(f^{-1}(B)\right)$$

    simply because it is not true.

    But the following is correct

    if $y\in B$ and $y\notin f(X)$ then $y\notin f\left(f^{-1}(B)\right)$

    and that is sufficient to prove that, in general, the inclusion is proper.

    on the other hand, if $B\subseteq f(X)$ then you will not be able to find such a $y$ and $B=f\left(f^{-1}(B)\right)$
    Thanks from detalosi
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. mapping
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Apr 10th 2011, 12:14 AM
  2. Complex open mapping & conformal mapping problems.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Feb 22nd 2011, 07:26 AM
  3. Mapping onto
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Aug 30th 2009, 05:28 PM
  4. Mapping: one-to-one and onto
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: Nov 6th 2008, 07:41 PM
  5. Help in Mapping..
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Jul 2nd 2008, 07:13 PM

/mathhelpforum @mathhelpforum