Thread: 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!

The easy way is to shift a countable set.
Let .
Then define

Originally Posted by Plato

The easy way is to shift a countable set.
Let .
Then define

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

? It's clearly injective, and maps to the interval ...

not ...

Originally Posted by sunmalus

not ...

Well, if by you mean the open set ( without the endpoints) then that is what that function does. It's just notation. There is no need for the function to be surjective.

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.