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March 13th, 2017, 07:05 PM   #21
SDK
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Quote:
Originally Posted by zylo View Post
Prove it or reference.

1/n is a member of (0,1) for every n.
1/n converges. That's the important point. The fact that 0 doesn't belong to (0,1) is irrelevant.
f(1/n) converges.
As usual, I think you are just trolling. In the off chance you aren't, let me reiterate. The concepts of homeomorphism between topological spaces and whether or not those spaces are complete could not have less to do with one another. The moment you start talking about convergence of a homeomorphism you have already taken a wrong turn.

Inverse tangent is a homeomorphism by the following proof:

1. It is continuous
2. It is a bijection between $(-\frac{\pi}{2},\frac{\pi}{2})$ and $\mathbb{R}$
3. Its inverse is $\tan x$ which is also continuous.

This is all that is required. Notice I never talked about convergence of anything, because homeomorphism is not concerned with this at all.

Last edited by skipjack; March 13th, 2017 at 11:18 PM.
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March 13th, 2017, 09:27 PM   #22
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<personal rambling>

Quote:
Originally Posted by SDK View Post
As usual I think you are just trolling.
For my part I can never know the intention behind a handle. I enjoy the challenge of writing clear and compelling responses. The other day after giving someone hints on their triangle inequality problem I finally just gave up and posted a proof. They never came back.

I would rather interact with someone responsive than waste my time doing someone's homework.

When I get frustrated I try to remember that seven billion people managed to not feel compelled to reply. If I feel compelled to reply it must be something about me, not about some anonymous handle on the Internet.

The Turing test tells you more about yourself than it does about the other person. That's what Turing never understood.

</personal rambling>

Last edited by Maschke; March 13th, 2017 at 09:42 PM.
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March 13th, 2017, 09:49 PM   #23
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Quote:
Originally Posted by zylo View Post
Guff
That just proves that you never take in anything of substance that other people write. I don't know whether you aren't capable of understanding or whether you are just too stubborn to admit that you are wrong, but I don't really care.

The reference in question is this one: Map (0,1) to R

Where Maschke showed you chapter and verse of your text where it clearly states that your sequence $\{\frac1n\}$ on $(0,1)$ DOES NOT CONVERGE no matter what rubbish you post to the contrary.
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March 14th, 2017, 12:51 PM   #24
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If I'm trolling, I'm not very successful at it.

Nothing so far contributes to or detracts from OP.

The homeomorphism of tan in post #26 is simply a statement of what is given, not a proof.


EDIT1
-pi/2 + 1/n is in (-pi/2,pi/2) and converges.
tan(-pi/2 + 1/n) converges to tan(-pi/2), which doesn't exist.

EDIT2
In anticipation of Maschke's objection:
1/n is a convergent sequence in R+.
1/n converges to 0, which is not in R+.
Ever hear of a boundary point?

Last edited by zylo; March 14th, 2017 at 01:27 PM.
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March 14th, 2017, 01:28 PM   #25
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Quote:
Originally Posted by zylo View Post
The homeomorphism of tan in post #26 is simply a statement of what is given, not a proof.
tan = sin/cos. Sin and cos are linear combinations of the complex exponential which is continuous. So tan is continuous except when cos is 0, namely +/- pi/2. You can Google the standard trig proof.

If I had to do arctan from first principles I'd be in trouble but you can look it up. Maybe show tan is an open map (sends open sets to open sets) which in conjunction with it being a bijection would show that its inverse is continuous.

There's also a definition of arctan in terms of complex log.

Quote:
Originally Posted by zylo View Post
-pi/2 + 1/n is in (-pi/2,pi/2) and converges.
Not in (-pi/2,pi/2). But there's a limit (haha I see what I did there!) to how many times people can tell you the same thing. Rudin already clarified this point. Convergence depends on the sequence AND on the ambient space.

Last edited by Maschke; March 14th, 2017 at 01:40 PM.
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March 14th, 2017, 01:56 PM   #26
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Quote:
Originally Posted by zylo View Post
1/n converges to 0, which is not in R+.
Ever hear of a boundary point?
You are talking out of your backside again. Instead of pretending that you know everything, READ YOUR BOOK.

Since the concept of "convergence" is something that depends purely on a definition, we may ask why that definition has been chosen. The answer is, presumably, that it has been chosen to make further theorems more easily and naturally expressible. However, since those further theorems (that you are trying to make use of) are based on the definition in your book, you have to use that definition. If you decide to use a different definition, the theorems you are trying to use no longer apply without re-writing them to be based on that different definition.

Last edited by v8archie; March 14th, 2017 at 02:01 PM.
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March 15th, 2017, 07:04 AM   #27
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tan is a bijection but not a homeomorphism between (-pi/2,pi/2) and R.

tan is a bijection and homeomorphism between [-pi/2,pi/2] and Re.

EDIT
"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."
https://en.wikipedia.org/wiki/Complete_metric_space

In other words, you can have convergent sequences in M that don't have limits in M, ie, boundary points.

Last edited by zylo; March 15th, 2017 at 07:15 AM.
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March 15th, 2017, 08:59 AM   #28
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Quote:
Originally Posted by zylo View Post
tan is a bijection but not a homeomorphism between (-pi/2,pi/2) and R.

tan is a bijection and homeomorphism between [-pi/2,pi/2] and Re.

EDIT
"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."
https://en.wikipedia.org/wiki/Complete_metric_space

In other words, you can have convergent sequences in M that don't have limits in M, ie, boundary points.
The Wiki page is worded badly. Stick to Rudin, not Wiki.
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March 17th, 2017, 08:14 AM   #29
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Is Tan homeomorphic from (0,1) to R? No.

In R, Lim 1/n = 0 but 1/n never equals 0.
In R, Lim Tan 1/n = Tan (Lim 1/n) = Tan 0,
Tan$\displaystyle ^{-1}$Tan 0 = 0

Tan=romsek's tan
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March 17th, 2017, 08:48 AM   #30
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That's not the mapping that was suggested though, is it? $\tan x$ doesn't map $(0,1)$ to $\mathbb R$. If you want a homeomorphism $(0,1) \mapsto \mathbb R$ you use $$\tan{\bigg(\big(x-\tfrac12\big)\pi\bigg)}$$

Under that mapping, $x_n = \frac1n$ doesn't converge in $(0,1)$ or, under the mapping, in $\mathbb R$.

Last edited by v8archie; March 17th, 2017 at 08:51 AM.
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