March 21st, 2017, 01:02 PM | #1 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 | Is 1/x a homeomrphism from x>0 to y>0?
Is 1/x a homeomrphism from x>0 to y>0? No. $\displaystyle y = \lim_{x\rightarrow \infty}1/x = 0.$ R is complete. Definition a) A function f: (M1) →(M2) 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. This definition implies M1 and M2 are metric spaces (distance defined). |
March 21st, 2017, 01:05 PM | #2 |
Senior Member Joined: Aug 2012 Posts: 1,972 Thanks: 550 |
Good God Almighty why are you starting yet another thread pushing your same incorrect idea? How on earth can x > 0 and y > 0 NOT be homeomorphic? They're the same exact set described using a different variable. Would you falsely claim that the identity function on the positive reals is not a homeomorphism? After all if f(x) = x is the identity function, then $\displaystyle \lim_{x \to 0} x = 0$, which is then subject to your same objection. So I ask you: Is the identity function a homeomorphism from the positive reals to the positive reals? This all comes down to the "puzzler" I tossed out a while back. Continuous functions do not necessarily preserve completeness. When you grok this you will be enlightened. Last edited by Maschke; March 21st, 2017 at 01:22 PM. |
March 21st, 2017, 01:29 PM | #3 |
Senior Member Joined: Sep 2015 From: USA Posts: 2,040 Thanks: 1063 | |
March 21st, 2017, 01:46 PM | #4 |
Senior Member Joined: Aug 2012 Posts: 1,972 Thanks: 550 |
It occurs to me that there's a better way to explain this. What the t-shirt should really say is: Continuous functions don't necessarily preserve Cauchy sequences. So the sequence $(\frac{1}{n})$ is Cauchy, and does not attain a limit in $(0,1)$, showing that $(0,1)$ is not complete. If we apply the continuous function $f(x) = \frac{1}{x}$ to that sequence, we get the sequence $1, 2, 3, 4, ...$ which does not converge in the reals. Yet the reals are complete, so what happened? Well, $1, 2, 3, 4, \dots$ is not Cauchy and therefore its non-convergence doesn't count against the completeness of the reals. Last edited by Maschke; March 21st, 2017 at 01:51 PM. |
March 21st, 2017, 02:39 PM | #5 |
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra |
I suspect that he refuses to accept that a Cauchy sequence doesn't necessarily converge in a given domain. In other words, he claims that $\{\frac1n\}$ converges in $(0,\infty)$. This then causes problems down the line. As usual, if you don't accept the standard definitions, you can't use the theorems that use those definitions. |
March 21st, 2017, 06:34 PM | #6 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 | Quote:
$\displaystyle \lim_{x\rightarrow a}1/x$ $\displaystyle \epsilon$ y>0 for every a in x>0, in which case answer to OP is yes. From that point of view, tan is homeomorphic on (-$\displaystyle \pi$/2,$\displaystyle \pi$/2). Don't recall anyone making that, or any other, rational argument, for or against OP. So 1/x did serve a purpose, to distill a question to its more transparent essence. Many thanks to all participants, whose responses I found quite motivational. So what is answer to OP? Yes or No. It reminds me of a real number being defined by a decimal for all n, or lim as n $\displaystyle \rightarrow \infty$. | |
March 21st, 2017, 06:51 PM | #7 | |
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra | Quote:
Decimals are defined for both, but this concept of a sequence being convergent only if its limit exists in the domain does relate to the idea of sequences diverging to $+\infty$ or $-\infty$. Such sequences are quite different in character from other divergent sequences such as $\sin{(n)}$ and the difference is highlighted by the existence of such mappings under which a Cauchy sequence can be mapped to a sequence that diverges to $+\infty$ or $-\infty$. Last edited by greg1313; March 22nd, 2017 at 08:00 AM. | |
March 21st, 2017, 07:36 PM | #8 |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 | |
March 21st, 2017, 08:47 PM | #9 |
Senior Member Joined: Aug 2012 Posts: 1,972 Thanks: 550 | Not in the positive reals. I don't even know what is this "0" you talk about. I live in the universe of the positive integers, and we don't have any 0 in our universe. The positive reals are a set. You know how set theory works. Everything we know about a set depends on its members. There are no "secret numbers" clinging to the outside of a set like barnacles. Last edited by Maschke; March 21st, 2017 at 09:00 PM. |
March 22nd, 2017, 03:44 AM | #10 | |
Math Team Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra | Quote:
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