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

• Oct 1st 2010, 06:31 AM
sunmalus
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?
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).

• Oct 1st 2010, 07:03 AM
Plato
The easy way is to shift a countable set.
Let $\displaystyle T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}$.
Then define $\displaystyle f(x) = \left\{ {\begin{array}{r,l} {\frac{1}{{x^{-1} + 2}},} & {x \in T\setminus\{0\}} \\ {x,} & {x \in [0,1]\backslash T} \\ \end{array} } \right.~\&~f(0)=\frac{1}{2}$
• Oct 1st 2010, 08:15 AM
Swlabr
Quote:

Originally Posted by Plato
The easy way is to shift a countable set.
Let $\displaystyle T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}$.
Then define $\displaystyle f(x) = \left\{ {\begin{array}{r,l} {\frac{1}{{x^{-1} + 2}},} & {x \in T\setminus\{0\}} \\ {x,} & {x \in [0,1]\backslash T} \\ \end{array} } \right.~\&~f(0)=\frac{1}{2}$

Can you not just divide everything by 2 and add a quarter,

$\displaystyle f(x) = \frac{2x+1}{4}$? It's clearly injective, and maps to the interval $\displaystyle [\frac{1}{4}, \frac{3}{4}]\subset (0, 1)$...
• Oct 1st 2010, 08:28 AM
sunmalus
$\displaystyle$f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$$not \displaystyle f(x) \colon [ 0,1 ] \longrightarrow [\frac{1}{4},\frac{3}{4}]... • Oct 1st 2010, 08:33 AM Swlabr Quote: Originally Posted by sunmalus \displaystyle f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$$ not $\displaystyle f(x) \colon [ 0,1 ] \longrightarrow [\frac{1}{4},\frac{3}{4}]$...

Well, if by $\displaystyle ]0, 1[$ you mean the open set $\displaystyle (0, 1)$ ($\displaystyle [0, 1]$ without the endpoints) then that is what that function does. It's just notation. There is no need for the function to be surjective.
• Oct 1st 2010, 09:08 AM
sunmalus
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.