# [bi|sur|in]jective coordinates

Discussion in 'Physics & Math' started by RJBeery, Sep 29, 2010.

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1. ### Guest254Valued Senior Member

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What is confusing you?

3. ### RJBeeryNatural PhilosopherValued Senior Member

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x --> y
Give the values of y for the following values of x: 0, .1, .5, .999999, 1

5. ### Guest254Valued Senior Member

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I'll assume you meant to say please! Do you see that:

$0 \notin \left{ 1, \frac{1}{2}, \frac{1}{3}, \ldots \right\}$

So can you see what f(0) is? Do you see that

$0\cdot 1 = \frac{1}{10} \in \left\{ 1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{10}, \frac{1}{11}, \ldots \right\}$

So can you see what f(1/10) is? Do you still need our help, or is this clear enough?

7. ### RJBeeryNatural PhilosopherValued Senior Member

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Yes, (I think) I see what you're doing I just don't understand why it's necessary. So...

0 --> 0
.1 = 1/10 --> 1/11
.5 = 1/2 --> 1/3
.999999 --> .999999
1 --> 1/2

Is that right? I presume we're doing this so that the inverse mapping gives a unique answer as well, I just need to understand why that's the case. Seriously Guest, I don't mean to come across as dense (or rude), I got completely RIPPED last night but I still thought in my hung-over stupor I'd be able to grasp this. I just need to STFU until my brain recovers

8. ### Guest254Valued Senior Member

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I'm delighted to be able to say I've taught you something! Of course, you're welcome.

Well you've certainly changed your tune, regarding how beneficial you find our exchanges. What a fickle character you must be.

So now I've done all the work for you, it should be very straight forward for you to provide us with a bijection between the sphere and the plane. I shall give you another HUGE hint: use the bijection I have given you to map the sphere to the sphere minus a single point.

9. ### rpennerFully WiredValued Senior Member

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The inverse mapping is a function because where the mapping is not an identity, it maps X (which is a subset of the interval) to X.

Last edited: Oct 1, 2010
10. ### RJBeeryNatural PhilosopherValued Senior Member

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Guest: Hey bro if admitting that I was wrong about something (...ANYTHING, apparently) and putting up with your smug condescension during a lesson that I suspect doesn't even pertain to my OP is going to put us on good terms, then I'm happy to oblige!

I'm serious though, I need to recover before I try understanding this further...

11. ### Guest254Valued Senior Member

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And now we're brothers! As I mentioned, what a fickle character you must be.

Disappointed you couldn't finish off the exercise though.

12. ### QuarkHeadRemedial Math StudentValued Senior Member

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1,686
Maybe we should cut straight to the chase.

Unless I am mistaken, and I frequently am, what Guest is driving at is something like this:

Every mathematical object we are ever likely to encounter can be described as a set with some additional properties (needless to say, sets don't have additional properties)

So let's say that our "mathematical object", say a group, a vector space, a topological space,... or whatever, is a set with the additional properties mandated by the axioms of the space we are working in. Usually, these additional properties involve some sort of binary operation (I am reasonably sure this should be an "always" but I am happy to be corrected)

Call $|X|$ as the underlying set for our object $X$ which has these additional properties.

In general, one requires that the maps $f: X \to Y$, as structured sets, preserve these properties, in which case they are called "morphisms" between these so-called structured sets.

But it remains the case, as Guest has tried to convince all here, that there are set functions $g:|X| \to |Y|$ (recall these are the underlying unstructured sets) that need not respect the structure of our space.

Sorry if I have derailed this thread even further

Last edited by a moderator: Oct 3, 2010
13. ### Guest254Valued Senior Member

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No, that's not what I'm saying, at all. And how you've inferred that, I shall never know. But well done: in a thread where someone is struggling with something so basic as a coordinate chart, you've managed to bring up

Topological spaces, compactness, Heine-Borel, topological definition of continuity...

and you've managed to finish off with a post with category-theoretic flavour. Good work.

14. ### RJBeeryNatural PhilosopherValued Senior Member

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OK, I think I've internalized what you did, Guest. I'll explain my way, as I would likely mangle the hell out of the formal description:
If we wanted to map the infinite integer set A {0..inf} to the infinite integer set B {1..inf} it would be trivial. f(x) --> x+1, and the inverse f(y) --> y-1. In a sense, we're "pushing the train" here...

Now even though the sets [0..1] and [0..1) also only differ by a single member, such a function isn't available because, well, the real numbers aren't countable. What we need to do, in this case, is select a subset from the reals that ARE countable (and distinct), which will allow us to "push the train", and then let f(x) = x for the remaining member mappings. What you did was choose any real that is representable in the form of x = 1/n, where n is an integer.

Now that we know this, we can expand this method to an infinite number of such mappings. Another example would be:

For any x such that x = 1/2^n, f(x) --> 1/2^(n+1)
For all other x, f(x) = x

Another:
For x = 1, f(x) --> 1/2
For x such that x = 1/pi(n), f(x) --> 1/pi(n+1) [where pi(n) = the nth prime number]
For all other x, f(x) = x

This is very interesting and I thank you for it. I don't believe this helps me on my OP though because this whole business of "if x conforms to such-and-such then use this function" is just another way of using multiple mappings, in my opinion. How is this any different than saying:

"While studying black holes, for x not on EH use Schwarzschild coordinates, otherwise use the following 2-sphere function...?"

15. ### Guest254Valued Senior Member

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1,056
Surely, you can't still be confused by this? I thought earlier you could "see" how this easy lemma could provide you with a bijection between the sphere and the plane!

If you find you're lacking in the creative-thinking-department, feel free to ask for more help.

16. ### przyksquishyValued Senior Member

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You may not think it helps you, but it satisfies literally what you originally asked for. Using Guest's example, give me any "coordinate" in [0,1) and I can tell you the unique point in [0,1] it represents, or vice-versa. Give me any "coordinate" in the plane, and I can tell you the unique point on the sphere it represents. Give me any point on the sphere, and I can tell you its unique "coordinate" in the plane. That's what it means to have a bijective mapping.

By your own definition, you can also use the real number line as a coordinate chart on the entire plane. For example, one way is to write out the decimal representation of a real number (your "coordinate"), and separate out the even and odd digits. So for instance the coordinate 123.456 might represent the point (13.5, 2.46) on the plane. This is a surjective[sup]*[/sup] mapping, which satisfies the requirement you gave in the OP. But is this really the sort of thing you would want to call a coordinate mapping?

[sup]*[/sup]In principle it's possible to construct bijections between $\mathbb{R}$ and $\mathbb{R}^{2}$ (they have the same cardinality). The example I gave isn't one because some real numbers have multiple decimal representations (eg. the infamous 0.999... = 1), and there are pairs of coordinates such as 1.5050505... and 0.5959595... which represent the same point on the plane.

17. ### Green DestinyBannedBanned

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why did they use the notation [,) and [,], why not just [,] and [,]? Is it to destinguish which coordinate system you are using from another?

18. ### arfa branecall me arfValued Senior Member

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Aha, a question I know the answer to.

The [,] notation means a closed interval, (,) means an open interval, and [,) sort of means closed at the left, open at the right. It's just another way to define inequalities, for instance all points < r, for r the radius of a circle, is an open set, but all points <= r is a closed set. Whether you want to include the point (0,0) is a matter of definition of intervals, then.

19. ### Green DestinyBannedBanned

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Ah, I see. Thank you.

20. ### Green DestinyBannedBanned

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You know RJ, if you are failing to understand what they say, that's not truely a failing of your behalf, but rather those who are teaching you.

21. ### RJBeeryNatural PhilosopherValued Senior Member

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OK that was just lame...if we're going to be brothers I need to teach you how to deliver zingers.
Very possible, but I also asked for:
So, even after I said more than once that I suspected this wasn't what I was looking for, we stayed on the path. I, because I was hoping I was wrong; and Guest, because he's not used to getting attention beyond being AN's backup singer (see bro? that's how you do it! :thumbsup

Here is something else I've asked for:
So your point is that the answer is no, because anything can be mapped bijectively? How would one apply this to mapping the "infinite observer A + local infalling observer B + sufficiently large black hole + event horizon" bijectively?

Last edited: Oct 4, 2010
22. ### RJBeeryNatural PhilosopherValued Senior Member

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I would normally agree with that but I don't think I've misunderstood anything posted yet. It just took me a bit to understand was was going on "under the hood" of Guest's post which is a far cry from simply seeing the method and copying it.

23. ### Green DestinyBannedBanned

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Okey dokey. I haven't followed the thread well, just bits here and there