Results 1 to 6 of 6

Math Help - f: [0,1]-->]0,1[, f injective

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    f: [0,1]-->]0,1[, f injective

    Hello,
    I'm looking for a injective function defined as f: [0,1]-->]0,1[.
    How should I start?
    I'm trying to start with what it implies:
    f:A--->B
    A=[0,1]={x:0<=x<=1}
    B=]0,1[={f(x):0<f(x)<1}
    f(A)=B
    x=y => f(x)=f(y) for all x,y in A.
    It's not arctan(x) because im(arctan(x)) with x in[0,1] is not ]0,1[. So I rather think that f is only injective in [0,1].
    I'm also thinking maybe I need to find f(x)^-1 first and then f(x), (since f is injective).

    Thanks in advance!
    Last edited by sunmalus; October 1st 2010 at 07:02 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,661
    Thanks
    1616
    Awards
    1
    The easy way is to shift a countable set.
    Let T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}.
    Then define <br />
f(x) = \left\{ {\begin{array}{r,l}<br />
   {\frac{1}{{x^{-1} + 2}},} & {x \in T\setminus\{0\}}  \\<br />
   {x,} & {x \in [0,1]\backslash T}  \\ \end{array} } \right.~\&~f(0)=\frac{1}{2}
    Last edited by Plato; October 1st 2010 at 07:27 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Plato View Post
    The easy way is to shift a countable set.
    Let T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}.
    Then define <br />
f(x) = \left\{ {\begin{array}{r,l}<br />
   {\frac{1}{{x^{-1} + 2}},} & {x \in T\setminus\{0\}}  \\<br />
   {x,} & {x \in [0,1]\backslash T}  \\ \end{array} } \right.~\&~f(0)=\frac{1}{2}
    Can you not just divide everything by 2 and add a quarter,

    f(x) = \frac{2x+1}{4}? It's clearly injective, and maps to the interval [\frac{1}{4}, \frac{3}{4}]\subset (0, 1)...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    62
     $f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$ not f(x) \colon [ 0,1 ] \longrightarrow [\frac{1}{4},\frac{3}{4}]...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by sunmalus View Post
     $f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$ not f(x) \colon [ 0,1 ] \longrightarrow [\frac{1}{4},\frac{3}{4}]...
    Well, if by ]0, 1[ you mean the open set (0, 1) ( [0, 1] without the endpoints) then that is what that function does. It's just notation. There is no need for the function to be surjective.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2009
    Posts
    62
    I am very sorry but I meant f bijective and not f injective. So the example given by Plato still works because it is also surjective.
    My apologies again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Injective
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 1st 2010, 03:15 AM
  2. If gof is injective, then f is injective
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 17th 2009, 11:10 PM
  3. B & AB injective but A not injective
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 24th 2009, 10:33 PM
  4. g(f(x) is injective, is f injective?
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: November 9th 2008, 07:23 PM
  5. Injective
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 23rd 2006, 11:09 AM

Search Tags


/mathhelpforum @mathhelpforum