My Math Forum Is 1/x a homeomrphism from x>0 to y>0?

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March 23rd, 2017, 07:10 AM   #11
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Quote:
 Originally Posted by Maschke 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.
1/x is not continuous on the reals.

March 23rd, 2017, 07:47 AM   #12
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 Originally Posted by Maschke So I ask you: Is the identity function a homeomorphism from the positive reals to the positive reals?
No.

$\displaystyle \lim_{x\rightarrow \infty}x=\infty$, which doesn't exist.

 March 23rd, 2017, 08:01 AM #13 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81 Is 1/x a homeomorphism from x>0 to y>0? 1) No. $\displaystyle y = \lim_{x\rightarrow \infty}1/x = 0,$ 0 $\displaystyle \not\epsilon$ y>0 2) Yes. $\displaystyle \lim_{x\rightarrow a}1/x =1/a$ $\displaystyle \epsilon$ y>0 for every a in x>0 It's a legitimate question.
 March 23rd, 2017, 10:46 AM #14 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81 For f(x) =1/x from x>0 to y>0 $\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}$ doesn't exist, but not relevant. For f(x) = x on x>0 $\displaystyle \lim_{x\rightarrow 0}x=0$, but not relevant. homeomorphism requires continuity for every x of x>0. If this is point you were trying to make, I accept it; but not relevant to OP since I was questioning the other end.
March 23rd, 2017, 11:25 AM   #15
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Quote:
 Originally Posted by zylo 1/x is not continuous on the reals.
It's continuous on the positive reals though, which was the domain in question.

 March 23rd, 2017, 01:23 PM #16 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,006 Thanks: 81 $\displaystyle \lim_{x\rightarrow a}1/x =1/a$ for all a in x>0. a=$\displaystyle \infty$ is not a number. In general, if f homeomorphic from A to B, $\displaystyle \lim_{x\rightarrow a}f(x) = f(a)$ $\displaystyle \epsilon$ B for all a in A. a=$\displaystyle \infty$ is not a number. That simple, logical, explanation resolves all issues, including tan.
 March 23rd, 2017, 03:24 PM #17 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,685 Thanks: 2175 Math Focus: Mainly analysis and algebra Yes, all convergent sequences in A map to convergent sequences in B. That's been the idea right from the start, hasn't it?
March 23rd, 2017, 03:26 PM   #18
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 Originally Posted by Maschke 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.
This is back to the idea that the metric gets distorted under the transformation because Cauchy-ness is defined in terms of the metric.

March 27th, 2017, 07:44 PM   #19
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Quote:
 Originally Posted by zylo Is 1/x a homeomrphism from x>0 to y>0?
Yes

e^{-x} is a homeomorphism from x$\displaystyle \geq$0 to 0<y$\displaystyle \leq$1

Strictly by definition of H. Completeness or limits at infinity have nothing to do with this.

e$\displaystyle ^{-x}$ continuous for all x$\displaystyle \geq$0 makes sense, e$\displaystyle ^{-x}$ continuous at $\displaystyle \infty$ doesn't make sense:
$\displaystyle \lim_{x\rightarrow \infty}f(x)= f(?)$

 Tags 1 or x, homeomrphism, x>0, x>0, y>0, y>0