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I've got an idea about how to show this but I'm not sure.

I have a few functions that are 1-to-1 and onto for this:

$f(x)$ from $(a,b)$ to $(c,\infty)$ will be: $f(x) :=$ $[(b-a)$ $*$ $c]$$/(b-x)$.
$g(x)$ from $(a,b)$ to $(-\infty,d)$ will be: $f(x)$ $:=$ $[(a-b)$ $*$ $d]$$/$$(a-x)$.
not sure about how to get the $(-\infty,\infty)$ thing.

Now, since any of these functions send an open neighborhood to an open neighborhood, that means they are open and continuous, am I right?

mihir
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CIsForCookies
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    Although your functions seem fine, let me suggest you a nicer function: $\tan x$. It sends $(0, \frac{\pi}{2})$ to $(0, \infty )$ and maps $( - \frac{\pi}{2}, \frac{\pi}{2} )$ to $( - \infty , \infty )$. – ThePortakal Sep 07 '15 at 15:42
  • Ok. Yours is vet nice, but my question remains: does that prove homomorphism? – CIsForCookies Sep 07 '15 at 15:52
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    If you also want a nice function from $(0,\infty)\rightarrow(-\infty,\infty)$, $\ln x$ works nicely. Or more generally, $(c,\infty)\rightarrow(-\infty,\infty)$ with $\ln(x-c)$. – mjh Sep 07 '15 at 17:11

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To prove it's a homeomorphism, it is enough to proceed as you have, which will prove it is a continuous open bijection, and so a homeomorphism. As for showing the homeomorphism to the whole real line, it might be easier to prove it starting from one of the two rays, and using a similar idea to the approach you've already taken.

Alternately, you can adapt any of the fine answers to this question.

Community
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Cameron Buie
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  • Just to be sure, the statement "any of these functions send open neighborhood to an open neighbourhood" is valid? No need of proof? – CIsForCookies Sep 07 '15 at 17:30
  • Well, you do need to prove it. It shouldn't be too hard to prove that such functions map open subintervals to open subintervals, though. – Cameron Buie Sep 07 '15 at 17:33
  • Still it might be preferable to proceed by showing that the functions' respective inverses are continuous. – Cameron Buie Sep 07 '15 at 17:35
  • I'm not sure how. am I allowed to use metrical proof using epsilon and delta? – CIsForCookies Sep 07 '15 at 17:39
  • That certainly works. Or, you could prove that the functions and their inverses map open subintervals to open subintervals, so the functions are open and continuous. – Cameron Buie Sep 07 '15 at 17:59
  • You just need to pick the method that's easiest for you to manage. It might be easiest to prove that all rational functions are continuous on their domains of definition, then prove that any real-valued monotone continuous function on a generalized real interval (that is: the empty set, a singleton, a ray, the whole real line, or one of the usual intervals) of $\Bbb R$ is an open map of the interval onto its image (which will again be a generalized real interval). – Cameron Buie Sep 07 '15 at 18:04
  • Even if that approach is a bit harder, it's still good practice, and gives you a handy result for later! – Cameron Buie Sep 07 '15 at 18:05
  • I'm so confused. I thought I'm not allowed using metric proof because this is a topological space, and if so I'm at a loss. If using metric proof is allowed, I think I can just say that an open ball (x,r) is mapped to open ball(f(x), f(r)), and that's it – CIsForCookies Sep 07 '15 at 18:11
  • Sure, it's a topological space, but the topology is induced by the metric, so using the metric should be fine. However, if you're concerned about it, bear in mind that the $\epsilon$-$delta$ style of proof can be translated into terms of preimages of open sets. – Cameron Buie Sep 07 '15 at 18:15
  • How can you tell it's included by the metric? maybe those sets indicated in the question were not open sets. Maybe the question is using the trivial topology. Can that be? – CIsForCookies Sep 07 '15 at 18:21
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    Typically, one assumes the usual topology on the real line, unless it's specified otherwise. As for how to see that the usual topology is induced by the metric, note that the open ball centered at $x$ of radius $r$ is $(x-r,x+r),$ which is a basis element of the usual topology. On the other hand, for $a<b,$ the interval $(a,b)$ is the open ball centered at $\frac{a+b}2$ of radius $\frac{b-a}2.$ – Cameron Buie Sep 07 '15 at 18:31
  • OK, now I got it, thx a lot :) – CIsForCookies Sep 07 '15 at 19:23