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

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 March 11th, 2017, 11:24 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 Homeomorphic mapping from (a,b) to R Definitions: a) A function f: (a,b) → R is homeomorphic (bi-continuous) if it has the following properties: f is a bijection (one-to-one 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 one-to-one 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 one-to-one (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 12:07 PM. Reason: correct f^-1
March 11th, 2017, 11:41 AM   #2
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Quote:
 Originally Posted by zylo Definitions: a) A function f: (a,b) → R is homeomorphic (bi-continuous) if it has the following properties: f is a bijection (one-to-one 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.
What is S? And if S is an arbitrary metric space, aren't you just giving a tautologous definition of convergence? A convergent sequence converges?

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:
 Originally Posted by zylo Theorem: The bounded open interval (a, b) is not homeomorphic to the real numbers R for any a < b.
Doesn't the tangent function map $(-\frac{\pi}{2}, ~ \frac{\pi}{2})$ continuously and bijectively to the reals? And isn't its inverse the arctangent also continuous?

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 12:15 PM.

 March 11th, 2017, 12:13 PM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 S is a metric space. Sorry, I thought that was obvious from the context (R).
March 11th, 2017, 12:31 PM   #4
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Quote:
 Originally Posted by zylo S is a metric space. Sorry, I thought that was obvious from the context (R).
Do you understand that (b) is not the correct definition of completeness?

March 11th, 2017, 12:50 PM   #5
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 Originally Posted by zylo 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).
Which of the following do you not agree with?
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, 01:07 PM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 "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 01:20 PM.
March 11th, 2017, 01:50 PM   #7
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 Originally Posted by zylo If a sequence sn in (a,b) converges, then f(sn) converges if f is homeomorphic. sn=(a+1/n) converges.
That's true of any continuous function, bijective or not. Continuous functions preserve limits. That's often taken as the definition in calculus.

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 01:57 PM.

March 13th, 2017, 07:55 AM   #8
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Quote:
 Originally Posted by Maschke 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?
1/n converges to 0, which is not a positive real number. It is also Cauchy convergent (every convergent sequence is Cauchy convergent). But it is not Cauchy convergent in the space of positive real numbers.

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, 08:31 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 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 $\pm\infty$. Thanks from Maschke Last edited by v8archie; March 13th, 2017 at 08:33 AM.
 March 13th, 2017, 11:38 AM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,363 Thanks: 100 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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