March 13th, 2017, 12:00 PM  #11 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 Math Focus: Dynamical systems, analytic function theory, numerics 
I'm having trouble even determining if you are asking a question somewhere. However, I have a few comments which have already been pointed out but should be emphasized. 1. The word homeomorphic describes topological spaces, not functions. Thus your first definition makes no sense since a function can't be homeomorphic. I think what you mean to say is the following: Two spaces $A,B$ are homeomorphic if there exists a continuous bijection, $f\colon A \to B$ whose inverse is also continuous. In this case, the function $f$ is called a homeomorphism. 2. You are incorrect that $(a,b)$ is not homeomorphic to $\mathbb{R}$. The problem with your supposed proof by contradiction is that you are presuming completeness is a topological invariant (preserved under homeomorphism) and it isn't. In fact, its strictly worse than it appears. It isn't even true that the continuous image of a Cauchy sequence is Cauchy. You should ask yourself why you think completeness should be preserved as this might help you understand something more clearly. 
March 13th, 2017, 01:24 PM  #12  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra  Quote:
However, the range of $\arctan{(x)}$ from the extended real domain is not the open interval $(\frac{\pi}2,\frac{\pi}2)$, it is the closed interval $[\frac{\pi}2,\frac{\pi}2]$. So your functions no aren't inverses.  
March 13th, 2017, 01:30 PM  #13 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
"..a homeomorphism or topological isomorphism or bi continuous function is a continuous function between topological spaces that has a continuous inverse function." https://en.wikipedia.org/wiki/Homeomorphism A Euclidean space is a topological space. Frankly, I used homeomorphic as a shortcut for a long definition: f is 1:1 from every point of A to every point of B and continuous in both directions. "You are incorrect that (a,b) is not homeomorphic to R." I have proved that (a,b) is not homeomorphic to R in OP. I have defined my terms carefully and given a simple, transparent, logical proof. What is wrong with the proof? 
March 13th, 2017, 02:10 PM  #14 
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364  
March 13th, 2017, 03:24 PM  #15  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
1/n is a member of (0,1) for every n. 1/n converges. That's the important point. The fact that 0 doesn't belong to (0,1) is irrelevant. f(1/n) converges.  
March 13th, 2017, 03:28 PM  #16 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra 
Why bother with reference books if you are going to ignore every definition that you don't like? For reference, your second and third statements were wrong before, they are wrong now and they will remain wrong until you decide to pay attention to your text and think logically about what you are saying. I'm not getting into another of these circular arguments with you. If you can't be bothered to put in the tiny amount of effort required to understand the point, that's your lookout. I have an "ignore" button. Last edited by v8archie; March 13th, 2017 at 03:33 PM. 
March 13th, 2017, 04:01 PM  #17 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  
March 13th, 2017, 04:52 PM  #18 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,941 Thanks: 2267 Math Focus: Mainly analysis and algebra 
What's wrong with the one from your own book that Maschke gave you?

March 13th, 2017, 05:09 PM  #19  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88  Quote:
To which v8archie replied: Where does Rudin prove "tan/arctan are homeomorphisms from an open interval to the reals."?  
March 13th, 2017, 05:36 PM  #20  
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364  Quote:
As a straight line through the origin sweeps out an angle in $(\frac{\pi}{2}, ~ \frac{\pi}{2})$, its slope hits every point in $(\infty, ~ \infty)$ and vice versa. You don't even need trigonometry for this, it's a fact taught in analytic geometry. Open sets are mapped to open sets in both directions. Last edited by Maschke; March 13th, 2017 at 05:38 PM.  

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