March 17th, 2017, 10:10 AM  #31  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91  Quote:
Quote:
The Principle: If f is continuous in R (assumed), Lim f(sn) exists if sn converges in R.  
March 17th, 2017, 10:18 AM  #32 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  
March 17th, 2017, 03:29 PM  #33 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
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, 03:55 PM  #34  
Senior Member Joined: Aug 2012 Posts: 1,661 Thanks: 427  Quote:
$\displaystyle \lim_{x \to \infty} \tan^{1} x = \frac{\pi}{2}$. Why are we arguing about high school trig? Last edited by Maschke; March 17th, 2017 at 03:59 PM.  
March 17th, 2017, 05:35 PM  #35 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  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, 07:23 AM  #36 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
Assume f homeomorphic from (a,b) to R, a<b. For any $\displaystyle \delta$, y=f(x) is bounded on (a+$\displaystyle \delta$, b$\displaystyle \delta$) $\displaystyle \rightarrow$ $\displaystyle \lim_{y\rightarrow \infty}f^{1}(y) = a$ and $\displaystyle \lim_{y\rightarrow \infty}f^{1}(y) = b$ because f$\displaystyle ^{1}$ continuous on R. $\displaystyle \rightarrow$ f is not homeomorphic. Example: y = tanx is not homeomorphic from ($\displaystyle \pi $/2,$\displaystyle \pi $/2) to R. 
March 18th, 2017, 07:39 AM  #37 
Global Moderator Joined: Dec 2006 Posts: 18,247 Thanks: 1439 
$\displaystyle \lim_{y\rightarrow \infty}f^{1}(y) = a$ 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, 02:37 PM  #38 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
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, 04:30 PM  #39 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 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, 11:20 PM  #40 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
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=(ba)/$\displaystyle \pi$, 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$\displaystyle \pi$(x1/2) is a homeomorphism from (0,1) to R 

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