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I am constructing a bijection from open interval $(0,1)$ to open interval $(a,b)$ for $a,b\in\mathbb{R}$ and $a<b$.

I tested the line through points $(0,1)$ and $(a,b)$, namely $f(x)=1+\frac{b-1}{a-0}(x-0)$.

But say $(a,b)=(3,4)$ giving $f(x)=1+\frac{4-1}{3-0}(x-0)$ or $f(x)=1+x$, then we see that $0<x<1$ gives $1<f(x)<2$ not $3<f(x)<4$.

Instead, this seems to map intervals $(0,3)$ to $(1,4)$ or $(0,a)$ to $(1,b)$.

Why does this line not map interval $(0,1)$ to interval $(a,b)$ ?

In this answer the bijection is $(-1,1)$ to $(a,b)$, and the line through the points suffices: https://math.stackexchange.com/a/2876242/1098426

I am avoiding wordiness, but if you have questions, please let me know.

The problem I am doing regards a continuous function with uncountably many points greater than some value, but that is merely my motive (MSE requires I state this) and is unrelated.

isaac
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    Think about the line between the points $(0,a)$ and $(1,b)$. – Anne Bauval Mar 05 '24 at 16:28
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    You want the line through the points $(0, a)$ and $(1, b)$. – Robert Shore Mar 05 '24 at 16:28
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    @AnneBauval Thank you. I tried this, but I see now, after your suggestion, and after retrying, that my first try was incorrect. I appreciate you taking the time to help. – isaac Mar 05 '24 at 16:32
  • @AnneBauval This appears to answer my question in a more general case. Perhaps my question is duplicate, in this sense. All the same, I will close my question with an answer, and provide a brief remark on the max and min. Thank you again. – isaac Mar 05 '24 at 16:49
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    Better put your answer in the linked post than in your "abstract duplicate". Anyway, you won't "close your question". – Anne Bauval Mar 05 '24 at 16:51
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    @AnneBauval Ah, I took my duplicate to be more concrete, but I will leave it then and spend my time on more important things. Like a broken record, I do appreciate you helping me. – isaac Mar 05 '24 at 16:54
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    The "stretching ratio" needs to be "length of new interval"$\div$ "length of old interval = $\frac {b-a}{1-0}$. And the "offset of $x$ in the original interval" is $(x-0)$. And the "new starting point" is $a$. So the formula of "the new output of $x$" = "new starting point" + "new offset of the output" = "new starting point" + "stretching ratio" $\times$ "offset of $x$ in the original interval" is $f(x)=a + \frac {b-a}{1-0}=(x-0)$. You just got the initial formula horked. – fleablood Mar 05 '24 at 17:17
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    Argh. Typo. I mean $f(x) = a+\frac {b-a}{1-0}(x-0)$. – fleablood Mar 05 '24 at 17:23
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    Hmmm.... you know. I always thought of these mappings as "stretching" and "shifting". I never really thought of trying to put $(0, 1)$ on the $x$ axis and $(a,b)$ on the $y$ axis and ploting the line between $(0,a)$ and $(1,b)$. That would work just as well. Just don't make tho mistake of confusing $(0,1)$ as a point and $(a,b)$ as a point and making a line between those. That line is not at all relevant to the issue. – fleablood Mar 05 '24 at 17:40

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