Sets: Prove that (0,1) ~ (1, 2) using bijection 2x + 1

[tashe]

I have this example:
Prove that :
(0,1)~(1,3) which is in my lesson for cardinal numbers.
I have this answer:
We define a bijection: f0,1)~(1,3) with f(x) = 2x+1

Does anyone knows how this was achieved?
thanks

 

[pka]

Re: Cardinal number

\(\displaystyle f(a) = f(b) \Rightarrow \quad 2a + 1 = 2b + 1 \Rightarrow \quad a = b\).
Does that mean that \(\displaystyle f\) is one-to-one?

\(\displaystyle f^{ - 1} (x) = \frac{{x - 1}}{2}\) so if \(\displaystyle s \in (1,3)\) and \(\displaystyle t = f^{ - 1} (s)\)
the what is \(\displaystyle f (t)\)?
What does that show?

 

[tashe]

Re: Set

I have no idea how its solved. But I did some guess work, and according to that f(x) is one-to-one. I think the answer is to find a function which maps every number from the first set to a number in the second set, one-to-one, (bijection) ,with the sets being with different cardinal number.
I just have no idea how they found that f(x)=2x+1 is the answer.(which is correct)

 

[pka]

Re: Set

I have no idea how its solved.
I just have no idea how they found that f(x)=2x+1 is the answer.(which is correct)

Please do not take offence at my remarks because none is intended.
I ask you: “Are you ready for this level of question”?
The point is: there is nothing special about that particular function.
It just happens to work. There are several others that work equally as well.

To prove that two sets are equipotent, one finds a bijection, any bijection, between the two sets

 

[tashe]

Re: Set

No problem man. If you keep explaining me you can easily reach the 4000th post .
So as a sum up, the point is to find a bijection in this case 2x+1. Do you know the way it is found?
Thats my final question. I wont bother you anymore.
Respect

 

[pka]

Re: Set

Do you know the way it is found? Thats my final question.

There really a pedestrian answer for that question: it is the simplest bijection that works. Did you notice that a linear function will map open sets to open sets?