[bi|sur|in]jective coordinates

And now you ask for my help again! You're giving me mixed messages.

 

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Mixed messages? Actually, Guest, I don't believe I've ever asked for your help. Quite the opposite in fact. I said I'm giving you the benefit of the doubt here, though, because you wrote something that I didn't know and want to understand.

 

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Oh, so you don't want my help now? Which is it? Would you like me to explain to you how to construct a bijection from the sphere to the plane, or not?

Because I'm feeling charitable, I shall start you with a hint. Can you produce a bijection between [0,1] and [0,1)?

 

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Hey bro I know you're excited that I'm asking you a question but bring the drama down a notch. :thumbsup:

As for the bijection, err, nothing comes to mind but I can "see" the analogy to the 2-sphere. AlphaNumeric had a pretty good post about "eliminating the points" at the poles to make the 2-sphere a plane and then finding a second coordinate map to cover those points. But, if you're headed in a direction that requires multiple, overlapping coordinate maps you aren't adding anything to my knowledge or the thread, as I asked for a single coordinate system to map Reality.

 

Now, now - you shouldn't bite the hand that feeds you.

How hard have you thought about it? This would be a chance to do some out-of-the-box type thinking, so maybe you need to give it another go. All you need to do is construct a bijection between the set \( \{ x: 0 \leq x \leq 1\}\) and the set \(\{ x: 0 \leq x < 1\}\).

If you give up, that's fair enough - I'll give you the answer and you can no doubt "see" how to use it to construct a bijection from the sphere to the plane.

EDIT: This HUGE hint will hopefully give you the answer. Consider the set X={1, 1/2, 1/3, 1/4, ...}.

 

Another hint: a bijection is just a mapping that is invertible, nothing said about continuity.

Maybe this one is easier: Find a bijection between the interval (0,1) and the square (0,1)x(0,1).

 

OK so it's something like represent x as a fraction and...add 1 to the denominator...does that work?
1 --> 1/2
1/2 --> 1/3
999/1000 --> 999/1001
100/100 --> 100/101

Ah, though, it isn't one-to-one. How would this map y back to x distinctly for x = 1?

This kind of stuff isn't my strong point.

 

Well at least you tried. I was hoping that my hint would make it particularly easy to see the bijection:

\( f(x) = x \quad \textrm{for}\,\, x\notin X \quad \textrm{and} \quad f\left( \frac{1}{n}\right) = \frac{1}{n+1}\quad \textrm{for}\,\, n\in \{ 1,2,3, \ldots \}\)

But now I've given you this, I'm sure you can see it's trivial to produce a bijection between the sphere and the plane.

 

The bijective mapping of \([0,1]\) between \((0,1)\) is usually described as:

\(X|0 \le x \le 1=[0,1]\)

\(X|0 < x < 1=(0,1)\)

The mapping is then usually taken as

\(0 \rightarrow 1/2\)
\(1 \rightarrow 1/3\)
\(1/2 \rightarrow 1/4\)
\(1/3 \rightarrow 1/5\)

 

Nice example! I always presumed that there is no explicit example like this one and the only examples are the ones that can be constructed by the iteration in the proof of the Cantor-Bernstein theorem.

 

It is only because now I have seen the example, but you can take any infinite sequence of distinct numbers in (0,1) and do the same.

 

I invite everyone to google "bijective mapping of [0,1] between (0,1)" and note the first link:

http://answers.yahoo.com/question/index?qid=20090626053554AAuIYCd

Shame on you, Green Destiny.

 

Don't get you. I didn't copy it if that is what you mean.

In fact, it doesn't matter.

 

On the same page, it has basically what you wrote:

If n is a positive integer, then f(1/n) = 1/(n + 2)
Otherwise (that is, if x is not zero and cannot be written in the form 1/n for n a positive integer), then f(x) = x.


We can all play games like that.


Fact is you can't change these things - that was the answer full stop.

 

Of course you did! You were even daft enough to employ the exact same notation! But please, do carry on. I like you.

No, the map on that page produces a bijection between (0,1) and [0,1]. Mine produces a bijection between [0,1] and [0,1). But please, keep digging!

 

Well, actually I just did. This is a very nice page on bijective mappings.

http://www.staff.ncl.ac.uk/peter.jorgensen/MAS2211/MAS2211_25-48.pdf

 

Let \(I\) be an set with cardinality equal to \(\mathbb{N}\) or greater. Let \(g\) be a bijection between \(\mathbb{N}\) and a subset (not necessarily proper) of \(I\). Then one may construct a bijection between \(I\) and \(I \setminus \left{ g(0), ... , g(n) \right}\) by mapping \(x\) in the image of \(g\) to \(g(1+n+g^{-1}(x))\) augmented by the identity when \(x\) is not in the image of \(g\).

So for our present case with the clue given, we have:

\(f(x) = \left\{ { \frac{1}{\frac{1}{x}+1} \quad \textrm{if} \quad x \in X \\ x \quad \textrm{otherwise}} \right. \\ f^{-1}(x) = \left\{ { \frac{1}{\frac{1}{x}-1} \quad \textrm{if} \quad x \in X \setminus \{ 1 \} \\ x \quad \textrm{otherwise}} \right. \\ f^{-1}(f(x)) = \left\{ { x \quad \textrm{if} \quad x \in X \\ x \quad \textrm{otherwise}} \right. \\ f(f^{-1}(x)) = \left\{ { x \quad \textrm{if} \quad x \in X \setminus \{ 1 \} \\ x \quad \textrm{otherwise}} \right. \)

Lemma:
\( x \neq \frac{1}{a} \Rightarrow \frac{1}{\frac{1}{\quad \frac{1}{\frac{1}{x}-a} \quad }+a} = x \)

 

This seems pretty close to what I suggested, but how does this cover, for example, x=.75?

 

Well since \(\frac{3}{4} \not\in X\), then \(f(\frac{3}{4}) = \frac{3}{4}\).

I know I'm just an American, with precious little grasp of math and English, but are you sure that you are even trying?

 

rpenner, never before have I considered you to be one of "those guys"...oh well

Are we still talking about the same thing? Maybe I'm having a notation problem here but I need help understanding the answer...