1. cardinaty

I have no idea how to do this question.

Let a,b be real numbers with a<b. Prove that (a,b) has same cardinality as (-1,1)

I guess this is trying to find existance of bijection function between (a,b) and (-1,1), right? But I have no idea how to prove in details. Please help me. Thanks a lot.

2. As just giving you the answer for the bijection won't be good, try to come up with it. What is the simplest function you can think of that will also be one-to-one?

3. Originally Posted by Vlasev
As just giving you the answer for the bijection won't be good, try to come up with it. What is the simplest function you can think of that will also be one-to-one?
Thank you. First apologize for my typing, I'm glad you know I mean "Cardinality", not "Cardinaty", I'm an idiot, sorry.
The simple one to one I can think of would be y=x, is that good enough?

4. Originally Posted by tsang
Let a,b be real numbers with a<b. Prove that (a,b) has same cardinaty as (-1,1)
existance of bijection function between (a,b) and (-1,1), right? But I have no idea how to prove in details.
Try $\dfrac{2(x-a)}{b-a}-1$.

5. Haha no problem. I totally missed the typo myself.

As for your suggestion, that's very close. Notice what happens for y = x when you plug in the end points. You get y(a) = a and y(b) = b. But instead you want y(a) = -1 and y(b) = 1.

6. Originally Posted by Vlasev
Haha no problem. I totally missed the typo myself.

As for your suggestion, that's very close. Notice what happens for y = x when you plug in the end points. You get y(a) = a and y(b) = b. But instead you want y(a) = -1 and y(b) = 1.

Thanks a lot. I think I have got what you mean.
However, as this is the specific case, how can I generalize it, so I can make it become a proper "proof"? I feel I cannot just simply try to give examples.
Sorry keep asking you for help, I've just started Real Analysis course two weeks ago, so I haven't built strong foundation so far.

7. It won't be clear that you "have got what you mean" until you actually write down the function. Yes, there is a linear function that will map (-1, 1) "one to one and onto" the interval (a, b). What is it? And what do you mean by "specific case"? (a, b) is a very general interval. You could, I guess, generalize to "(a, b) has the same cardinality as (c, d)" for a, b, c, d four numbers such that a< b, c< d. What linear functions maps (c, d) onto (a, b)?

8. Sorry guys, but I'm having a lot of difficulty with this seemingly easy problem. I understand what the method of proof will be. We'll need to find some function with domain (a,b) that maps onto an interval (-1,1). We'll then need to show that this function is bijective, by proving both injection and surjection. My problem is that I simply cannot find such a function, that is both one-to-one and satisfies surjection (that is, for all values over (-1,1), there exists some x such that a<x<b).

I know I must be missing something obvious, I just don't see it.

9. Originally Posted by Ares_D1
Sorry guys, but I'm having a lot of difficulty with this seemingly easy problem. I understand what the method of proof will be. We'll need to find some function with domain (a,b) that maps onto an interval (-1,1). We'll then need to show that this function is bijective, by proving both injection and surjection. My problem is that I simply cannot find such a function, that is both one-to-one and satisfies surjection (that is, for all values over (-1,1), there exists some x such that a<x<b).

I know I must be missing something obvious, I just don't see it.

10. Originally Posted by Plato
Hmm, I didn't see that earlier, no. Will give it a shot and see how it goes. It does look very similar to a few of my earlier attempts.

EDIT: Fantastic, it works! For future reference though, was any important scratch work done to obtain this function, or just simple trial-and-error?

11. Any linear function is of the form f(x)= mx+ p where m and p are constants. If the function is to map (a, b) onto (-1, 1) then we must have f(a)= ma+ p= -1 and f(b)= mb+ p= 1. That gives you two equations to solve for m and p and they are particularly easy. If you subtract the first equation from the second you eliminate p: mb+ p- (ma+ p)= m(b- a)= 1- (-1)= 2.

m= 2/(b- a). Then the first equation becomes ma+ p= 2a/(b-a)+ p= -1 so that $p= -1- \frac{2a}{b-a}=$ $-\frac{b-a}{b-a}- \frac{2a}{b-a}= \frac{-b+ a- 2a}{b-a}= -\frac{a+ b}{b- a}$.

That gives $f(x)= mx+ p= \frac{2}{b- a}x-\frac{a+b}{b-a}= \frac{2x- (a+b)}{b-a}$.

For the more general problem I proposed, to map (a, b) to (c, d) (showing that any two intervals have the same cardinality), the equtions become ma+ p= c and mb+ p= d. Subtracting the first equation from the second, m(b- a)= d- c so that $m= \frac{d- c}{b- a}$, etc.

Linear functions are not only comparatively simple, but are always "one to one" and "onto". That's why everyone was suggesting you look for a linear function.