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Construct a bijection between $[1,2]$ and $[3,5)$.

So what I usually do if I pair the two outer number together to each other so like $3=1m+b$ and $5=2m+b$ and just solve for m and b but since 5 is not included, I am not sure how to construct it that makes it still a bijection. Thanks for any help!

Edit: this is not from $(0,1] \rightarrow (0,1)$ and also the brackets are different.

Asaf Karagila
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rickster38
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  • well this is from bracket bracket to bracket no bracket. Also, looking at it, I"m not sure how I would apply that to this problem. – rickster38 Nov 23 '15 at 06:53
  • To go the other way round, just take the inverse of the function. Also if you have a bijectiion $(1, 2] \to (3, 5)$, then it is easy to construct a bijection $[1, 2] \to [3, 5)$. –  Nov 23 '15 at 06:56
  • I"m not sure how you use the proposed answer. So I have to make a infinite sequence using (1,2]? – rickster38 Nov 23 '15 at 06:58
  • Do it step by step. Try to understand the argument (using infinite series) in the linked duplicate first. That is the hardest part. –  Nov 23 '15 at 07:02
  • Yeah, I"m not sure what this part means. "Choose an infinite sequence (xn)n⩾1(xn)n⩾1 of distinct elements of (0,1). " LIke do I make an inifinite sequences that is in between (0,1)? – rickster38 Nov 23 '15 at 07:07
  • Yes, something like $\frac 1n$ would do. –  Nov 23 '15 at 07:20

1 Answers1

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Definition .$N$ is the set of non-negative integers.....Take a strictly decreasing sequence $(x_n)_{n\in N}$ with $x_0=5$ and $\lim_{n\to \infty}x_n=3. $ Take a strictly decreasing sequence $(y_n)_{n\in N}$ with $y_0=2 $ and $\lim_{n\to \infty}y_n=1 . $ Let $f:[x_{n+1},x_n)\to (y_{n+1},y_n] $ be bijective for each $n\in N . $ And let $f(3)=1.$ (Note that $f$ has discontinuities.This is necessary.) For example let $x_n=3+2^{1-n}$ and $y_n=1+2^{1-n}$. And for $x\in [x_{n+1},x_n)$ let $f(x)=y_n+(y_{n+1}-y_n)(x_n-x_{n+1})^{-1}(x-x_{n+1}).$

DanielWainfleet
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  • What would be an example of a strictly decreasing sequence? like for the first on would it be like 5-2/n? Or what would be an example? – rickster38 Nov 23 '15 at 07:19
  • okay I think I understand it a little, but what would f(x) itself be? so like the first would be like $Xn = 5-4x^2/x^2$, but what f be itself – rickster38 Nov 23 '15 at 07:27
  • I added a specific example. – DanielWainfleet Nov 23 '15 at 07:30
  • wait on your f is from [a,b) to(a,b], mine is [a,b] to [a,b). Is it the same? – rickster38 Nov 23 '15 at 07:38
  • If $ f:A\to B$ is a bijection then the inverse function$ f^{-1} :B\to A $ is also a bijection. That is why you see the phrase "a bijection between $A$ and $B$". I just found it easier to construct it from [3,5) to [1,2] based on something related to this that I know. – DanielWainfleet Nov 23 '15 at 07:45
  • so then how would I take the inverse? how would I do that with your function ? – rickster38 Nov 23 '15 at 07:47
  • also what would be the best way to prove that it is a bijection – rickster38 Nov 23 '15 at 08:02
  • Observe that on the interval $[x_{n+1},x_n)$ , the function $f$, in my example, is linear so computing its inverse is elementary algebra . And $f^{-1}(1)=3$. – DanielWainfleet Nov 23 '15 at 08:05
  • Several question.For the interval, does it matter on what n it is on? Also I don't understand for thee function since we are use yn, wouldn't it matter what n was and not x, if we replaced everything is x replace n. Also I don't understand the inverse. – rickster38 Nov 23 '15 at 08:11
  • I meant for each and every n, computed separately for each n.. – DanielWainfleet Nov 23 '15 at 08:15
  • so would the inverse function be like $y=(x-y_n)/((y_(n+1)-y_n)(x_(n-x_(n+1)^-1)+x_(n+1)$ – rickster38 Nov 23 '15 at 08:20
  • can you just help me on what the inverse function would be – rickster38 Nov 23 '15 at 09:17