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∀w,x,y,z ∈ R, w < x and y < z. Given that information, supply a bijection between the two intervals. (w,x) and (y,z) Then after you find the bijection, provide a formal proof that what you found is a bijection.

(FACT: the interval (w,x) is defined as (w,x) = {p ∈ R|w < p < x})

The proof needs to use the definitions of injective and surjective, then it needs to formally show that the bijection that is found is a bijection.

user127662
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  • There is a somewhat opvious candidate: $t\mapsto \frac{t-w}{x-w}\cdot (z-y)+y$ – Hagen von Eitzen Apr 01 '14 at 17:23
  • Finding the bijection is an important step but the hard part is formulating a formal proof to prove that the bijection you found is infact a bijection. – user127662 Apr 01 '14 at 17:25
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    How formal do you need your formal proof to be? Which axiomatic framework do you work in? – hmakholm left over Monica Apr 01 '14 at 17:38
  • Good question, the proof will likely be around 6-12 line sentences. As far as formality goes, we can make assumptions about certain things. For instance we can assume that a bijection consists of a surjection and injection. The proof likely needs to derive from the injection and surjective definitions listed on this wikipedia page: http://en.wikipedia.org/wiki/Bijection,_injection_and_surjection We can also assume that we already know that an injection and surjection is proven true, so a lemma to prove these cases would not be necessary. – user127662 Apr 01 '14 at 17:45
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    In this case, I would much rather show that the given bijection has a two sided inverse. Are you allowed to assume that a function is a bijection if and only if it has a two sided inverse? If so, go with that. – Dan Rust Apr 01 '14 at 18:31

2 Answers2

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Define $f: (w,x) \rightarrow (y,z)$ by

$$f(\alpha):= y+ (z-y) \dfrac{\alpha -w}{x-w}.$$

I leave the details to you.

Frank
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(w,x)-->(0,x-w) can be achieved via a translation (bijection), call this f.

(y,z)-->(0,z-y) can be achieved via a translation (bijection), call this g.

Then (0,x-w)--->(0,z-y) can be achieved via a scaling (bijection), since we have $w\neq x$ and $z\neq y$ call this h.

Explicitly, we have: $f(t)=t-w$, $g(t)=t-y$, $h(t)=\frac{z-y}{x-w}\cdot t$, which are all bijective functions with inverses. Therefore, to get (w,x)-->(y,z), its just $g^{-1}hf$. (I don't know whether you are used to the archaic German notation or the modern one, mine is the modern one, meaning f first followed by h followed by the inverse of g)

The proof is simple, just show that each of these 3 functions are bijections, which is an easier feat (compared to the final function). Then use the fact that composition of 3 bijections are bijections.

If you have problem showing f, g, h are bijections, do ask me and I'll try to help.

user139068
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  • so in that problem, how do we actually go about proving those 3 functions are bijections? i know how with concrete examples but with abstract stuff like this I get confused. – user127662 Apr 02 '14 at 19:41
  • Ill do one. f:(w,x)->(0,x-w) ; t |--> t-w. Given any number r such that 0<r<w-x, one has f(r+w)=r and that x<r+w<w. This is surjectuvity – user139068 Apr 03 '14 at 04:30