My Math Forum Homeomorphic mapping from (a,b) to R

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March 17th, 2017, 09:10 AM   #31
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Quote:
 Originally Posted by zylo Is Tan homeomorphic from (0,1) to R? No. In R, Lim 1/n = 0 but 1/n never equals 0. In R, Lim Tan 1/n = Tan (Lim 1/n) = Tan 0, Tan$\displaystyle ^{-1}$Tan 0 = 0 Tan=romsek's tan

Quote:
 Originally Posted by v8archie That's not the mapping that was suggested though, is it? $\tan x$ doesn't map $(0,1)$ to $\mathbb R$. If you want a homeomorphism $(0,1) \mapsto \mathbb R$ you use $$\tan{\bigg(\big(x-\tfrac12\big)\pi\bigg)}$$ Under that mapping, $x_n = \frac1n$ doesn't converge in $(0,1)$ or, under the mapping, in $\mathbb R$.
I defined Tan as romsek's tan, which is what you gave. Nothing changes in the argument. I just used it to concentrate on the principle.

The Principle:

If f is continuous in R (assumed), Lim f(sn) exists if sn converges in R.

March 17th, 2017, 09:18 AM   #32
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Quote:
 Originally Posted by zylo I defined Tan as romsek's tan, which is what you gave. Nothing changes in the argument.
So the following is utter nonsense.
Quote:
 In R, Lim Tan 1/n = Tan (Lim 1/n) = Tan 0

 March 17th, 2017, 02:29 PM #33 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 969 Thanks: 78 Given $\displaystyle Tan^{-1}$ continuous on R. $\displaystyle \lim_{x\rightarrow -\infty}Tan^{-1}(x) = 0$ exists 0 not in (0,1) Tan not holeomorphic on (0,1) to R I will have to Revise OP proof accordingly.
March 17th, 2017, 02:55 PM   #34
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Quote:
 Originally Posted by zylo Given [MATH] $\displaystyle \lim_{x\rightarrow -\infty}Tan^{-1}(x) = 0$ exists 0 not in (0,1)
That's not true, just look at the graph of arctan.

$\displaystyle \lim_{x \to -\infty} \tan^{-1} x = -\frac{\pi}{2}$.

Why are we arguing about high school trig?

Last edited by Maschke; March 17th, 2017 at 02:59 PM.

March 17th, 2017, 04:35 PM   #35
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Quote:
 Originally Posted by zylo Given $\displaystyle Tan^{-1}$ continuous on R. $\displaystyle \lim_{x\rightarrow -\infty}Tan^{-1}(x) = 0$ exists 0 not in (0,1) Tan not holeomorphic on (0,1) to R I will have to Revise OP proof accordingly.
Instead of trying to be clever by writing $\DeclareMathOperator{\Tan}{Tan}\Tan{(x)}$, write it all down using the proper notation to demonstrate to yourself what a mess you are making of this. There is no value of $x \in \mathbb R$ for which $\Tan^{-1}{(x)}=0$.

 March 18th, 2017, 06:23 AM #36 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 969 Thanks: 78 Assume f homeomorphic from (a,b) to R, a
 March 18th, 2017, 06:39 AM #37 Global Moderator   Joined: Dec 2006 Posts: 16,783 Thanks: 1238 $\displaystyle \lim_{y\rightarrow -\infty}f^{-1}(y) = a$ doesn't imply that $f$ isn't homeomorphic. A function that's continuous on the open interval (a, b) needn't be bounded.
 March 23rd, 2017, 01:37 PM #38 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 969 Thanks: 78 In general, if f homeomorphic from A to B, $\displaystyle \lim_{x\rightarrow a}f(x) = f(a)$ $\displaystyle \epsilon$ B for all a in A. a=$\displaystyle \infty$ is not a number. Example: $\displaystyle \lim_{x\rightarrow a}1/x =1/a$ for all a in x>0. a=$\displaystyle \infty$ is not a number. That simple, logical, explanation resolves all issues, including tan (I agree, it's homeomorphic on (-pi/2,pi/2) Taken from: Is 1/x a homeomrphism from x>0 to y>0?
 March 23rd, 2017, 03:30 PM #39 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,548 Thanks: 2148 Math Focus: Mainly analysis and algebra As pointed out in the other thread, all you are saying is that convergent sequences in A map to convergent sequences in B, which is where we started.
 March 23rd, 2017, 10:20 PM #40 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 969 Thanks: 78 So tanx is a homeomorphiism from (-$\displaystyle \pi$/2, $\displaystyle \pi$/2) to R. Ref post #38. It follows there is a homeomorphism from (a,b) to R for any a < b: Let $\displaystyle \bar{x}$ = Ax+B $\displaystyle \bar{x}$ = a when x = -$\displaystyle \pi$/2 $\displaystyle \bar{x}$ = b when x = $\displaystyle \pi$/2 A=(b-a)/$\displaystyle \pi$, B=(a+b)/2 Example If a=0 and b=1, $\displaystyle \bar{x}$ = x/$\displaystyle \pi$ + 1/2, x = $\displaystyle \pi$($\displaystyle \bar{x}$ - 1/2), and tanx=tan$\displaystyle \pi$($\displaystyle \bar{x}$ - 1/2). Reverting to notation $\displaystyle \bar{x}$ = x f(x)= tan$\displaystyle \pi$(x-1/2) is a homeomorphism from (0,1) to R

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