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 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. Thanks from topsquark and v8archie Last edited by Maschke; March 21st, 2017 at 01:22 PM.
March 21st, 2017, 01:29 PM   #3
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 Originally Posted by Maschke Continuous functions do not necessarily preserve completeness.
needs to be on a t-shirt

 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. Thanks from v8archie 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
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 Originally Posted by zylo Is 1/x a homeomrphism from x>0 to y>0? No. $\displaystyle y = \lim_{x\rightarrow \infty}1/x = 0.$ R is complete.
But I can see another point of view:
$\displaystyle \lim_{x\rightarrow a}1/x$ $\displaystyle \epsilon$ 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
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 Originally Posted by zylo $\displaystyle \lim_{x\rightarrow a}1/x$ $\displaystyle \epsilon$ y>0 for every a in x>0
The point is that this is only true because $\displaystyle \lim_{x\rightarrow a} \tfrac1x = \tfrac1a \in (0,\infty)$. That is, the limit of the sequence exists in the domain on which we are working. If you have a sequence whose limit is not within the domain on which we are working, the sequence cannot converge whether it is Cauchy or not. This is precisely the point that has been made to over the course of all of these threads.

Quote:
 Originally Posted by zylo It reminds me of a real number being defined by a decimal for all n, or lim as n $\displaystyle \rightarrow \infty$.
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
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Quote:
 Originally Posted by zylo Is 1/x a homeomrphism from x>0 to y>0?
No: f(x$\displaystyle _{n}$) = 1/n is Cauchy convergent to y=0.

Yes of my previous post also holds.

March 21st, 2017, 08:47 PM   #9
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 Originally Posted by zylo No: f(x$\displaystyle _{n}$) = 1/n is Cauchy convergent to y=0.
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
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 Originally Posted by v8archie 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.
qed

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