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?

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- August 8th 2010, 03:21 AM #1

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## 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.

- August 8th 2010, 03:24 AM #2

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- August 8th 2010, 03:38 AM #3

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- August 8th 2010, 03:41 AM #4

- August 8th 2010, 03:51 AM #5

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- August 8th 2010, 04:11 AM #6

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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.

- August 8th 2010, 04:37 AM #7

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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)?

- August 8th 2010, 05:07 AM #8

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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.

- August 8th 2010, 05:41 AM #9

- August 8th 2010, 05:53 AM #10

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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?

- August 8th 2010, 08:12 AM #11

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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 .

That gives .

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 , 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.