March 11th, 2017, 12:24 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91  Homeomorphic mapping from (a,b) to R
Definitions: a) A function f: (a,b) → R is homeomorphic (bicontinuous) if it has the following properties: f is a bijection (onetoone and onto), f is continuous, the inverse function f$\displaystyle ^{1}$ is continuous.* b) Complete: Every convergent sequence in S converges to a member of S. Theorem: The bounded open interval (a, b) is not homeomorphic to the real numbers R for any a < b. Proof: 1) (a,b) is not complete. 2) R is complete 3) Assume f is homeomorphic from (a,b) to R. 4) a+1/n converges to a limit (which is not in (a,b)). 5) f(a+1/n) converges to a limit (which is in R because R is complete). > 6) a+1/n converges to a limit in (0,1). Contradiction. *Reference: https://en.wikipedia.org/wiki/Homeomorphism EDIT: In mathematics, a bijection, bijective function or onetoone correspondence is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set. There are no unpaired elements. In mathematical terms, a bijective function f: X → Y is a onetoone (injective) and onto (surjective) mapping of a set X to a set Y. From: https://en.wikipedia.org/wiki/Bijection Last edited by zylo; March 11th, 2017 at 01:07 PM. Reason: correct f^1 
March 11th, 2017, 12:41 PM  #2  
Senior Member Joined: Aug 2012 Posts: 1,661 Thanks: 427  Quote:
Isn't it part of the definition of convergence that the sequence has to converge to something in the same set? Isn't that what the Rudin definition says? Doesn't that Rudin page explicitly say that $\{\frac{1}{n}\}$ does NOT converge in $(0,1$)? Isn't $\{\frac{1}{n}\}$ a Cauchy sequence? Doesn't the existence of a Cauchy sequence that does NOT converge prove (by definition) that $(0,1)$ is not complete? Can you imagine infinitely "stretching out" an open interval to the entire real line continuously, like a strand of Turkish taffy? Do people still know what Turkish taffy is? Does it have something do with General Flynn? Quote:
Can not a high school student who understands nothing more than the "rise over run" definition of the slope of a straight line visualize this homeomorphism by considering a line through the origin that maps an angle in $(\frac{\pi}{2}, ~ \frac{\pi}{2})$ to a slope in $(\infty, ~ \infty)$, and vice versa? Conclusion: Homeomorphism does not preserve completeness! Completeness is a metric property but not a topological property. Last edited by Maschke; March 11th, 2017 at 01:15 PM.  
March 11th, 2017, 01:13 PM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
S is a metric space. Sorry, I thought that was obvious from the context (R).

March 11th, 2017, 01:31 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,661 Thanks: 427  
March 11th, 2017, 01:50 PM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  Quote:
1) $a+\frac1n$ does not converge in $(a,b)$ because $a \not \in (a,b)$; 2) $f(a+\frac1n)$ does not converge in $\mathbb R$ because $f(a+\frac1n) \to \pm\infty$? Both are true, but I suspect that you haven't realised either. 1) is by the reference Masche pointed you to in your own book. 2) I don't have a proof for, but essentially it is because $f$ is continuous and the boundary points of $(a,b)$ must therefore map to the boundary points of $\mathbb R$. Please don't go on another spree of nonsense threads all based on your failure to understand either of these two points.  
March 11th, 2017, 02:07 PM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
"In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M." From https://en.wikipedia.org/wiki/Complete_metric_space Cauchy Sequence: https://en.wikipedia.org/wiki/Cauchy_sequence A convergent sequence is Cauchy. I was trying to limit the jargon and definition overload and not distract from the point. Tan(pi /2,pi /2) to R is not homeomorphic. If it were, Lim tan(pi/2+n) would have to exist because Lim(pi/2+n) exists. Tan(pi/2) doesn't exist. See OP, which makes an end run around $\displaystyle \infty$ by sticking strictly to the definitions. EDIT: If a sequence sn in (a,b) converges, then f(sn) converges if f is homeomorphic. sn=(a+1/n) converges. Last edited by zylo; March 11th, 2017 at 02:20 PM. 
March 11th, 2017, 02:50 PM  #7  
Senior Member Joined: Aug 2012 Posts: 1,661 Thanks: 427  Quote:
When I showed you the passage in Rudin where he notes that $\{\frac{1}{n}\}$ does not converge in the positive reals, did you not understand what he's saying? Or do you believe he's wrong? He's very clear on this point. A Cauchy sequence is a sequence that should "morally" converge but that does not converge because the limit is not there. So when we imagine  and by we, I mean we ALL have this intuition  that $\{\frac{1}{n}\}$ "should" converge to $0$, the problem is that it DOESN"T converge, because $0 \notin (0,1)$. If we imagine we're stuck in the universe of $(0,1)$, there's no zero there. So there is nothing for the sequence $\{\frac{1}{n}\}$ to converge to. That's the cleverness of the definition of a Cauchy sequence. A Cauchy sequence is a sequence that should "morally" converge but may or may not actually converge. If EVERY Cauchy sequence converges, we call the space complete, because it actually contains all the moral limits. But if a space is missing a moral limit  that is, if there's a Cauchy sequence that doesn't converge  then it's not complete. There's a hole where there morally should be a point. And that  pardon the pun  is the point. A great example is the rational numbers. The sequence $1, 1.4, 1.41, 1.412, \dots$ should morally converge to the square root of $2$. But as Euclid showed there is no rational number whose square is $2$. So the rational numbers are not complete. The rational numbers have holes where we wish there were points. The real numbers are designed to plug up the holes in the rationals. Last edited by Maschke; March 11th, 2017 at 02:57 PM.  
March 13th, 2017, 08:55 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91  Quote:
https://en.wikipedia.org/wiki/Complete_metric_space In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M. Intuitively, a space is complete if there are no "points missing" from it (inside or at the boundary. I have been totally consistent with that definition. For example, allow the extended real number system, then range tanx is R. Assume tan is a homeomorphism between (pi/2,pi/2) and R. 1) Lim (pi/2 + 1/n)= pi/2, which does not belong to (pi/2,pi/2) 2) Lim tan(pi/2 + 1/n)=$\displaystyle \infty$, which belongs to R. 3) arctan $\displaystyle \infty$=pi/2, which belongs to (pi/2,pi/2). Contradiction.  
March 13th, 2017, 09:31 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 Math Focus: Mainly analysis and algebra  $\infty \not \in \mathbb R$ Why do you keep claiming that it is? $\mathbb R$ contains only numbers. $\infty$ is not a number. Also $\arctan {(\infty)} \ne \frac{\pi}{2}$ the arctangent function is not defined at $\pm\infty$. Last edited by v8archie; March 13th, 2017 at 09:33 AM. 
March 13th, 2017, 12:38 PM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
I have already pointed out that tan(pi/2) doesn't exist. $\displaystyle \infty$ exists as a symbol in the extended real number system: baby Rudin, Def 1.39). I used it for tan(pi/2) to create an example for OP. If the extended number system bothers you, ignore the example. I don't wish to go off on a tangent about "$\displaystyle \infty$" again. 

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