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I need help with finding a bijection between these two intervals: [a,b) and [a,b]. It is told that a,b are in R (real numbers). I know how to construct a bijection if a,b are integers, but i have no idea how to do it if they are real. Any ideas?

Edit: If anyone is still watching, can you tell me if I'm doing this right?

So the bijection will be f(x) = 1) a if x = a and 2) b/p-1 if x != a and x = b/p , where p is in R - {0}.

By taking the p from R - {0} and not from N, I think this could work for real numbers.

Because a < b, every number from a to b can be written as b/p. So the function moves every number upwards if I express myself that way, thus reaching the b.

uagamaga
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    Can you write down what you have for integers? – Prahlad Vaidyanathan Dec 29 '13 at 13:48
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    First thing is to search this website. Then you can see that it has been covered about a zillion times before. Then you read some of the answers and decide if you understand them or not. If you do, great! If you don't, you can post a question specifying the answer and what you don't understand there. – Asaf Karagila Dec 29 '13 at 13:50
  • See this for the basic idea(s). – David Mitra Dec 29 '13 at 13:51
  • i know all that, i've done all the (0,1) bijections and i have read other answers. The thing that i don't understand is how to find that bijection for REAL NUMBERS. – uagamaga Dec 29 '13 at 13:53
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    What's the difference? – Shai Deshe Dec 29 '13 at 13:56
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    Can you come up with a bijection between $[0,1]$ and $[a,b]$? Perhaps something linear? (Also, let this be a lesson to you: none of us can read your mind. If you don't tell us what you've done, we can only assume that you didn't do it.) – Asaf Karagila Dec 29 '13 at 13:56
  • You might see this – Ross Millikan Dec 29 '13 at 14:03
  • @Ross: The OP points out that the case of $(0,1)$ was done; but the problem is inferring the general case from that one. – Asaf Karagila Dec 29 '13 at 14:05
  • @AsafKaragila: I thought $0$ and $1$ were real numbers in this case, but I see your point. – Ross Millikan Dec 29 '13 at 14:07
  • If they were integers, you could do something like f(x) = 1) b/n-1 if x = b/n and 2)x if x != b/n , where n is a natural number. But with real numbers you cant do that. edit: Actually, it's wrong, but you see my point. With integers you can work like that. – uagamaga Dec 29 '13 at 14:09
  • Why can't you do that with integers? – rfauffar Dec 29 '13 at 14:17
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    A story about hotel room told by D.Hilbert will be a good hint. – C Weid Dec 29 '13 at 14:43
  • That's a good hint, but it is about integer numbers. You cant move the numbers on one branch to 2n and on the other to 2n+1 because you work with real numbers, which may be irrational. – uagamaga Dec 29 '13 at 14:48
  • Voting to reopen. It seems clear from the OPs comments that how to prove $[0,1)\simeq [0,1]$ (which is what the duplicates are about), but has trouble seeing that $[a,b)\simeq [a,b]$ for arbitrary real $a<b$ can be done by the same methods. – hmakholm left over Monica Dec 29 '13 at 17:45

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It looks like you're confusing yourself by thinking that just because $a$ and $b$ are arbitrary reals, you must not mention $\mathbb Z$ or $\mathbb N$ anywhere in your construction. In particular, when you want to allow your $p$ to be an arbitrary nonzero real, you get yourself into something that's not easy to repair.

Instead partition the numbers in your intervals $[a,b)$ and $[a,b]$ into two classes:

  • Numbers that can be written as $a+\frac{b-a}{n}$ for some $n\in\mathbb N$.
  • All other numbers.

$b$ itself is in the first class (take $n=1$). Now make your bijection be the identity on the second class and shift everything one position in the first.

Even though $b-a$ can be irrational, there's nothing that prevents you from dividing it by a sequence of integers. You get more irrationals out of that, but that doesn't matter for this purpose -- the only important thing is that you get an infinite sequence of different real numbers inside the interval that you can shift by one position.

hmakholm left over Monica
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