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

1. ## 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).

2. The easy way is to shift a countable set.
Let $T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}$.
Then define $
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}$

3. Originally Posted by Plato
The easy way is to shift a countable set.
Let $T=\{0\}\cup\{\frac{1}{n}:n\in\mathbb{Z}^+\}$.
Then define $
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,

$f(x) = \frac{2x+1}{4}$? It's clearly injective, and maps to the interval $[\frac{1}{4}, \frac{3}{4}]\subset (0, 1)$...

4. $f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$ not $f(x) \colon [ 0,1 ] \longrightarrow [\frac{1}{4},\frac{3}{4}]$...

5. Originally Posted by sunmalus
$f(x) \colon [ 0,1 ] \longrightarrow ]0,1[$ 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.

6. 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.