March 23rd, 2017, 07:10 AM  #11  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Quote:
 
March 23rd, 2017, 07:47 AM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  
March 23rd, 2017, 08:01 AM  #13 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 
Is 1/x a homeomorphism from x>0 to y>0? 1) No. $\displaystyle y = \lim_{x\rightarrow \infty}1/x = 0,$ 0 $\displaystyle \not\epsilon$ y>0 2) Yes. $\displaystyle \lim_{x\rightarrow a}1/x =1/a$ $\displaystyle \epsilon$ y>0 for every a in x>0 It's a legitimate question. 
March 23rd, 2017, 10:46 AM  #14 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 
For f(x) =1/x from x>0 to y>0 $\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}$ doesn't exist, but not relevant. For f(x) = x on x>0 $\displaystyle \lim_{x\rightarrow 0}x=0$, but not relevant. homeomorphism requires continuity for every x of x>0. If this is point you were trying to make, I accept it; but not relevant to OP since I was questioning the other end. 
March 23rd, 2017, 11:25 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  
March 23rd, 2017, 01:23 PM  #16 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 
$\displaystyle \lim_{x\rightarrow a}1/x =1/a$ for all a in x>0. a=$\displaystyle \infty$ is not a number. 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. That simple, logical, explanation resolves all issues, including tan. 
March 23rd, 2017, 03:24 PM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
Yes, all convergent sequences in A map to convergent sequences in B. That's been the idea right from the start, hasn't it?

March 23rd, 2017, 03:26 PM  #18 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  This is back to the idea that the metric gets distorted under the transformation because Cauchyness is defined in terms of the metric.

March 27th, 2017, 07:44 PM  #19 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Yes e^{x} is a homeomorphism from x$\displaystyle \geq$0 to 0<y$\displaystyle \leq$1 Strictly by definition of H. Completeness or limits at infinity have nothing to do with this. e$\displaystyle ^{x}$ continuous for all x$\displaystyle \geq$0 makes sense, e$\displaystyle ^{x}$ continuous at $\displaystyle \infty$ doesn't make sense: $\displaystyle \lim_{x\rightarrow \infty}f(x)= f(?)$ 