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March 23rd, 2017, 07:10 AM   #11
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Quote:
Originally Posted by Maschke View Post
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.
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March 23rd, 2017, 07:47 AM   #12
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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.
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March 23rd, 2017, 08:01 AM   #13
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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.
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March 23rd, 2017, 10:46 AM   #14
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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.
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March 23rd, 2017, 11:25 AM   #15
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Quote:
Originally Posted by zylo View Post
1/x is not continuous on the reals.
It's continuous on the positive reals though, which was the domain in question.
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March 23rd, 2017, 01:23 PM   #16
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$\displaystyle \lim_{x\rightarrow a}1/x =1/a$ 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.
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March 23rd, 2017, 03:24 PM   #17
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Yes, all convergent sequences in A map to convergent sequences in B. That's been the idea right from the start, hasn't it?
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March 23rd, 2017, 03:26 PM   #18
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Quote:
Originally Posted by Maschke View Post
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.
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March 27th, 2017, 07:44 PM   #19
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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:
$\displaystyle \lim_{x\rightarrow \infty}f(x)= f(?)$
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