My Math Forum Homeomorphic mapping from (a,b) to R

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 March 13th, 2017, 12:00 PM #11 Senior Member   Joined: Sep 2016 From: USA Posts: 384 Thanks: 208 Math Focus: Dynamical systems, analytic function theory, numerics I'm having trouble even determining if you are asking a question somewhere. However, I have a few comments which have already been pointed out but should be emphasized. 1. The word homeomorphic describes topological spaces, not functions. Thus your first definition makes no sense since a function can't be homeomorphic. I think what you mean to say is the following: Two spaces $A,B$ are homeomorphic if there exists a continuous bijection, $f\colon A \to B$ whose inverse is also continuous. In this case, the function $f$ is called a homeomorphism. 2. You are incorrect that $(a,b)$ is not homeomorphic to $\mathbb{R}$. The problem with your supposed proof by contradiction is that you are presuming completeness is a topological invariant (preserved under homeomorphism) and it isn't. In fact, its strictly worse than it appears. It isn't even true that the continuous image of a Cauchy sequence is Cauchy. You should ask yourself why you think completeness should be preserved as this might help you understand something more clearly. Thanks from Maschke
March 13th, 2017, 01:24 PM   #12
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 Originally Posted by zylo $\displaystyle \infty$ exists as a symbol in the extended real number system

However, the range of $\arctan{(x)}$ from the extended real domain is not the open interval $(-\frac{\pi}2,\frac{\pi}2)$, it is the closed interval $[-\frac{\pi}2,\frac{\pi}2]$. So your functions no aren't inverses.

 March 13th, 2017, 01:30 PM #13 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 "..a homeomorphism or topological isomorphism or bi continuous function is a continuous function between topological spaces that has a continuous inverse function." https://en.wikipedia.org/wiki/Homeomorphism A Euclidean space is a topological space. Frankly, I used homeomorphic as a shortcut for a long definition: f is 1:1 from every point of A to every point of B and continuous in both directions. "You are incorrect that (a,b) is not homeomorphic to R." I have proved that (a,b) is not homeomorphic to R in OP. I have defined my terms carefully and given a simple, transparent, logical proof. What is wrong with the proof?
March 13th, 2017, 02:10 PM   #14
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 Originally Posted by zylo What is wrong with the proof?
You have a wrong definition of convergence. And your proof is contradicted by the very well known fact that tan/arctan are homeomorphisms from an open interval to the reals.

March 13th, 2017, 03:24 PM   #15
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 Originally Posted by Maschke And your proof is contradicted by the very well known fact that tan/arctan are homeomorphisms from an open interval to the reals.
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.

 March 13th, 2017, 03:28 PM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Why bother with reference books if you are going to ignore every definition that you don't like? For reference, your second and third statements were wrong before, they are wrong now and they will remain wrong until you decide to pay attention to your text and think logically about what you are saying. I'm not getting into another of these circular arguments with you. If you can't be bothered to put in the tiny amount of effort required to understand the point, that's your lookout. I have an "ignore" button. Last edited by v8archie; March 13th, 2017 at 03:33 PM.
March 13th, 2017, 04:01 PM   #17
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 Originally Posted by v8archie Why bother with reference books if you are going to ignore every definition that you don't like? I have an "ignore" button.
So you can't prove your statement or give a reference.

 March 13th, 2017, 04:52 PM #18 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra What's wrong with the one from your own book that Maschke gave you?
March 13th, 2017, 05:09 PM   #19
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 Originally Posted by Maschke And your proof is contradicted by the very well known fact that tan/arctan are homeomorphisms from an open interval to the reals.
To which I replied: "So you can't prove your statement or give a reference."

To which v8archie replied:
Quote:
 Originally Posted by v8archie What's wrong with the one from your own book that Maschke gave you?
Where does Rudin prove "tan/arctan are homeomorphisms from an open interval to the reals."?

March 13th, 2017, 05:36 PM   #20
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 Originally Posted by zylo Where does Rudin prove "tan/arctan are homeomorphisms from an open interval to the reals."?
It's not in Rudin. It's well known from high school trigonometry, though they don't talk about bijections and homeomorphisms. I proved it a few posts back in one of the four threads on the same subject you currently have going. Someone else provided a proof as well in one of those threads. The multiplicity of threads makes it tedious to try to go back and find anything.

As a straight line through the origin sweeps out an angle in $(-\frac{\pi}{2}, ~ \frac{\pi}{2})$, its slope hits every point in $(-\infty, ~ \infty)$ and vice versa. You don't even need trigonometry for this, it's a fact taught in analytic geometry. Open sets are mapped to open sets in both directions.

Last edited by Maschke; March 13th, 2017 at 05:38 PM.

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