My Math Forum Homeomorphism from open set to closed set

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 March 13th, 2017, 12:15 PM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 Homeomorphism from open set to closed set Theorem: In E an open (partially open) set M1 is not homeomorphic to a closed set M2. Proof: 1) Let Sn be a convergent sequence to a boundary point P of M1 which is not in M1. 2) f(Sn) converges to a point Q in M2. 3) f$\displaystyle ^{-1}$(Q)=P which is not in M1. Contradiction. Definitions 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^{−1} is continuous. E: Euclidean space. I suspect the theorem could be framed in Topological space, but not my cup of tea.
March 13th, 2017, 12:59 PM   #2
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 Originally Posted by zylo Theorem: In E an open (partially open) set M1 is not homeomorphic to a closed set M2.
Counterexample:

* In $\mathbb R$ it's certainly the case that $\mathbb R$ is homeomorphic to $\mathbb R$. This is witnessed by the identity function on the reals, a bicontinuous bijection.

* $\mathbb R$ is an open set.

* $\mathbb R$ is a closed set.

Therefore there is at least one open set in $\mathbb R$ that is homeomorphic to a closed set. The empty set is another.

Your error here is the same one you keep making. The definition of convergence requires that the limit is a member of the set in question. $\{\frac{1}{n}\}$ does not converge in the positive reals, exactly as Rudin says.

And the reason he explicitly mentions this example is because it's confusing to students. Everyone is entitled to be confused about this for a while. If you would just carefully review Rudin's definition of convergence you would see that $\{\frac{1}{n}\}$ does not converge in the positive reals.

"1) Let Sn be a convergent sequence to a boundary point P of M1 which is not in M1." That is simply not the definition of convergence.

Last edited by Maschke; March 13th, 2017 at 01:12 PM.

 March 13th, 2017, 01:42 PM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 I claim 1/n converges to 0. If you don't understand or believe that, sorry, I don't know what language you are speaking. Where does Rudin say 1/n doesn't converge to 0?
March 13th, 2017, 02:04 PM   #4
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 Originally Posted by zylo I claim 1/n converges to 0. If you don't understand or believe that, sorry, I don't know what language you are speaking. Where does Rudin say 1/n doesn't converge to 0?
Rudin, Principles of Mathematical Analysis, Second Edition (I believe this is the version you're working from), Chapter 3, "Numerical Sequences and Series," page 41.

I quoted this section here: Map (0,1) to R

I repeat the relevant passage. This is directly from Rudin as you can verify for yourself.

Quote:
 Originally Posted by Rudin It might be well to point out that our definition of "convergent sequence" depends not only on $\{p_n\}$ but on $X$; for instance, the sequence $\{\frac{1}{n}\}$ converges in $\mathbb R^1$ (to $0$) but fails to converge in the set of all positive real numbers ...
I have now typed in this passage verbatim twice from a book that you yourself claim to be working from. You can verify for yourself that Rudin explicitly calls out this exact case.

 March 13th, 2017, 03:57 PM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 I never questioned what Rudin said. I am saying that 1/n converges to 0. (Do I really have to say in R? That's ridiculous.) For example: 1/n belongs to (0,1) for every n. 1/n converges to 0, which is not in (0,1), but that doesn't matter. If 1/n converges, f(1/n) converges in R. The point is that if 1/n converges f(1/n) converges. The fact that Lim 1/n is not in (0,1) doesn't matter. Are you saying there are no sequences in an open set which converge to a boundary point?
March 13th, 2017, 03:59 PM   #6
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 Originally Posted by zylo I never questioned what Rudin said. I am saying that 1/n converges to 0. (Do I really have to say in R?
Yes. Because as Rudin explicitly notes, 1/n does not converge in (0,1). And if you deny that you ARE contradicting what Rudin said; and thereby misunderstanding the nature of convergence and completeness.

Quote:
 Originally Posted by zylo That's ridiculous.)
Which is no objection to it being true. Relativity, quantum theory, and the fact that Donald Trump is president are all ridiculous but evidently true.

Perhaps a historical perspective is in order. When Newton explained gravity using calculus, he well understood that he did not know how to explain what dy/dx (in modern notation) was.

It took 200 more years to develop the proper logical rigor, resulting in epsilonics.

Is epsilonics perhaps ridiculous? Well it's difficult, and many generations of students can tell you so. But it's the best model we've got, and it's worth learning on its own terms. So if you say it's ridiculous, I wouldn't even bother to argue the point. But it's still important and still worth learning. On its own terms. Straight from Rudin for that matter.

Last edited by Maschke; March 13th, 2017 at 04:03 PM.

 March 13th, 2017, 04:17 PM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 Where did I say 1/n converges to 0 in (0,1), to use Rudin's language? I said, again and again and again, 1/n converges to 0 which is not in (0,1).
March 13th, 2017, 04:43 PM   #8
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 Originally Posted by zylo Where did I say 1/n converges to 0 in (0,1), to use Rudin's language? I said, again and again and again, 1/n converges to 0 which is not in (0,1).
Which means it does not converge in (0,1), the point you refuse to acknowledge. And since you are using your misunderstanding to "prove" things known to be false, the burden is on you to clarify your understanding.

March 13th, 2017, 05:14 PM   #9
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 Originally Posted by Maschke Which means it does not converge in (0,1), the point you refuse to acknowledge. And since you are using your misunderstanding to "prove" things known to be false, the burden is on you to clarify your understanding.
"In mathematics, a Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses."
https://en.wikipedia.org/wiki/Cauchy_sequence

"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

*..if every Cauchy sequence in M converges in M." ie, there can be Cauchy (convergent) sequences in M that do not converge in M.

1/n is a Cauchy sequence (convergent sequence) in M= (0,1) whose limit is not in M.

March 13th, 2017, 05:41 PM   #10
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 Originally Posted by zylo *..if every Cauchy sequence in M converges in M." ie, there can be Cauchy (convergent) sequences in M that do not converge in M. 1/n is a Cauchy sequence (convergent sequence) in M= (0,1) whose limit is not in M.
You're almost there. 1/n is a DIVERGENT Cauchy sequence in (0,1). That proves by definition that (0,1) is not complete.

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