My Math Forum Map (0,1) to R

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 March 7th, 2017, 10:30 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 Map (0,1) to R Can you map (0,1) to R bijectively and continuously both ways? How? Reference: Maschke, Post #1 Relatively Open and Closed Sets EDIT: Yes. Ex: Tangent on $\displaystyle (-\pi /2, \pi /2)$ real analysis - Is there a bijective map from $(0,1)$ to $\mathbb{R}$? - Mathematics Stack Exchange Questions of completeness irrelevant because we are not talking about closed and bounded spaces. Could have been discussed in original post once I understood question but it was irrelevant to thread. Thanks from greg1313 and Joppy Last edited by zylo; March 7th, 2017 at 10:53 AM.
 March 7th, 2017, 10:46 AM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 $f(x) = \tan\left(\pi\left(x-\dfrac 1 2\right)\right),~x \in (0,1)$ Thanks from zylo
March 7th, 2017, 11:39 AM   #3
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 Originally Posted by romsek $f(x) = \tan\left(\pi\left(x-\dfrac 1 2\right)\right),~x \in (0,1)$
(0,1) maps to R. f=$\displaystyle \infty$ maps to 1, which isn't in (0,1).

 March 7th, 2017, 12:07 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,970 Thanks: 2291 Math Focus: Mainly analysis and algebra $\infty \not \in \mathbb R$, so there's no problem. Also $f(x)\ne\infty$ for all $x \in (0,1)$. Last edited by v8archie; March 7th, 2017 at 12:13 PM.
March 7th, 2017, 02:52 PM   #5
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Quote:
 Originally Posted by romsek $f(x) = \tan\left(\pi\left(x-\dfrac 1 2\right)\right),~x \in (0,1)$
v8archie is correct. Extended number system is irrelevant here. I saw this on my way to coffee break and didn't feel like running back to edit.

(0,1) is not complete. f maps every convergent sequence in (0,1) to a convergent sequence in R. f also maps 1/n which is convergent to 0 which isn't in (0,1) to a convergent sequence in R.

R is complete. Why doesn't the inverse of f map every convergent sequence in R to a convergent sequence in (0,1)? It does. It maps every convergent sequence in R to a sequence which converges "in" (0,1), including 0 and 1, which aren't in (0,1).

Any paradox is due to the ambiguity in the use of "converges." 1/n "converges" to 0 but 0 isn't in (0,1) so it doesn't "converge in (0,1)"

If that doesn't explain what the following "tells us," can anyone explain what does?

"Now here is a puzzler for you. The metric space R is complete. Every Cauchy sequence converges.
Now using the tangent/arctangent we have a homeomorphism (bijection continuous in both directions) between R and the open unit interval (0,1). But (0,1) is NOT complete, since the Cauchy sequence (1/n) does not converge. What does this tell us?"

March 7th, 2017, 03:21 PM   #6
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 Originally Posted by zylo f maps every convergent sequence in (0,1) to a convergent sequence in R.
No it doesnt. $\{\frac1n\} \to 0$ maps to a sequence that diverges to $-\infty$. $\{1-\frac1n\} \to 1$ maps to a sequence that diverges to $\infty$.

March 7th, 2017, 03:43 PM   #7
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 Originally Posted by zylo R is complete. Why doesn't the inverse of f map every convergent sequence in R to a convergent sequence in (0,1)?
It does. Yet one space is complete and the other isn't. It's a puzzler. I presented it to you as a point of interest. I wasn't trying to be mean to you. Absolutely no intent to be insulting in any way. I'm sorry you took it that way but no insult was meant and I hope you understand that.

It's an interesting point. Continuous functions map convergent sequences to convergent sequences; and there's a bicontinuous bijection between the reals and the open unit interval; yet the reals are complete and the open unit interval isn't.

What is the conclusion to be drawn from that?

That's not an insult, it's a puzzler. There's a mathematical conclusion to be drawn and I was just throwing it out there for discussion. Let's leave it open for a little while. It's intended to stimulate discussion, not just give an answer. It's quite subtle.

A clue is to consider $(\frac{1}{n})$. As $n$ gets large, the terms of the sequence bunch up close together. But their image under the continuous mapping to the reals spreads out. The topology doesn't change. What changes?

Last edited by Maschke; March 7th, 2017 at 03:51 PM.

March 7th, 2017, 05:36 PM   #8
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 Originally Posted by Maschke What changes?
Being outside my area of expertise, I may be about to talk nonsense. But I would suggest that the metric doesn't survive the transformation. Or rather, we are not using the image of the metric after the transformation - we continue to use the original metric. This clearly has an impact on the convergence of sequences because convergence is defined in terms of the metric.

March 7th, 2017, 06:39 PM   #9
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 Originally Posted by v8archie Being outside my area of expertise, I may be about to talk nonsense. But I would suggest that the metric doesn't survive the transformation. Or rather, we are not using the image of the metric after the transformation - we continue to use the original metric. This clearly has an impact on the convergence of sequences because convergence is defined in terms of the metric.
I think that's a good way to put it. The way someone explained this to me is that completeness is a metric property and not a topological property. Homeomorphisms (bi-continuous bijections) preserve the topology. Open sets get mapped to open sets in both directions. But they do not preserve completeness.

And if you think about it, that makes sense. Completeness is defined as "every Cauchy sequence converges." But Cauchy sequences are defined in terms of the metric. There's no such thing as a Cauchy sequence in a topological space.

Or as you noted, the metric gets stretched out. Continuous functions preserve limits, but they don't preserve Cauchy-ness.

I wanted to mention that convergence is still preserved. The sequence $(\frac{1}{n})$ is not convergent. There is nothing in the open unit interval for it to converge to! It's no different than the sequence $1, 2, 3, 4, ...$ that also has no limit but isn't thought of as having a "moral" limit. We always want $(\frac{1}{n})$ to have a moral limit of $0$, but there is no $0$ in the universe of the open unit interval.

Is a closed set a set that contains all of its limit points? Well then the open unit interval $(0,1)$ is a closed set in $(0,1)$.

But it's not a closed set in the reals. That I believe is the question Zylo was originally asking about. The relativity of closed/open sets. It depends on the ambient space, and if there is one.

Last edited by Maschke; March 7th, 2017 at 06:44 PM.

March 8th, 2017, 03:00 AM   #10
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Quote:
 Originally Posted by Maschke It's an interesting point. Continuous functions map convergent sequences to convergent sequences; and there's a bicontinuous bijection between the reals and the open unit interval; yet the reals are complete and the open unit interval isn't. What is the conclusion to be drawn from that?
You missed the point. Every convergent sequence in R is mapped to a convergent sequence in (0,1) except the one which maps to the convergent sequence whose limit is 0, which is not in (0,1), but a border point which doesn't map to R.

The conclusion is not a puzzle. It is a trick question.

Last edited by zylo; March 8th, 2017 at 03:17 AM.

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