Marcos

Posted on Thursday, 06 November, 2003 - 05:01 pm:

There's some concept I don't understand regarding the number of coordinates required to describe a point in n-dimensional space. No doubt this is probably due to my inadequate knowledge of the subject.

Basically, as |

² | = |

| then can't we represent every (x,y) with a single (unique) real z?
Extending to

ⁿ , can we simply represent each (x₁ ,x₂ ,...,x_n ) by a single real x₀ ? (obtained from the bijective function)
Doesn't this invalidate the (at least intuitive) idea that a point in n-space requires at least n coordinates to be described?

Thanks,
Marcos

David Loeffler

Posted on Thursday, 06 November, 2003 - 05:08 pm:

It depends how nicely you want your 'description' to be.

² and

are isomorphic as sets, if we ignore any further structure, but they are not isomorphic as topological spaces, so you can't find a bijection

² < ->

which is continuous (in both directions).

David

Ian Short

Posted on Thursday, 06 November, 2003 - 05:10 pm:

Ha! Interesting idea. So you can describe all of LaTeX Image

with only LaTeX Image

s worth of points. But can you find a useful such description?

Ian

Zacky Choo

Posted on Thursday, 06 November, 2003 - 05:34 pm:

I am sure it works for

. In other words, yes, you can describe all of

ⁿ in

s worth of points.
So if you restrict yourself to the integers, you will be fine. I think it works for the rationals as well, but this is much harder to prove.

However, I think it fails for

. The reason is mainly because if you do concoct some sort of map, you will inevitably miss something in between. And it can be proven that there is always something in between.

Or, in technical terms, though both

and

are infinite,

is countable but

is uncountable

Marcos

Posted on Thursday, 06 November, 2003 - 05:47 pm:

Ah right!
So, topologically

ⁿ is (in some intuitive sense) larger than

? (ie. can't "deform" one into the other)

Could there be a case (ie. definitely one in which continuity weren't an issue

) where it may be useful to represent

ⁿ by

?

I think that the difference between representing

² by

and what is being discussed in this thread is that as

is countable so representing

² by

(say on a computer) would definitely not save space as every two numbers in the 'list' of the

representation would correspond to at most one member of

² . With the continuum (here I mean |

|, in case continuum means something else

) things seem different as you can't form 'lists' and intuitively I get this impression that you can crunch the entire

ⁿ into an interval [0,1] or something like that with some interesting consequences.

I think I even confused myself in that previous explanation! But, you have to admit it's a confusing issue, particularly for me

...

Thanks for the input David and Ian,
Marcos

Marcos

Posted on Thursday, 06 November, 2003 - 05:55 pm:

Sorry Zacky, I missed your post...
Are you saying that you can't represent

? I'm guessing this is obvious as there's no bijection between the two so no matter how interesting a function you come up with you'll miss something out (as you have said).

Another point: Is there an equivalent concept/theory like topology which works for discrete countable sets? As I see it, topology studies only uncountable sets (with certain other important properties)
So that we can say that in this situation

² isn't isomorphic to

? Although I'm not quite sure when two sets would be isomorphic in this 'discrete topology' unless in the trivial (boy I hate that word) case when they are identical sets...

Hrm, more food for thought!

Thanks,
Marcos

Dan Goodman

Posted on Thursday, 06 November, 2003 - 06:00 pm:

Marcos, yup - that's all correct. I can't think of any circumstances when it's useful to identify

R^{n}

and

R

off hand, although identifying

N^{n}

and

N

is often useful (in, for example, the theory of computability). You can certainly crunch

R^{N}

to the interval (0,1). First of all, identify

R

and

R^{n}

by interlacing decimal representations, then identify

R

with (0,1) by the map

x \to \tan ((2 x - 1) π / 2)

(which continuously identifies (0,1) with

R

Dan Goodman

Posted on Thursday, 06 November, 2003 - 06:07 pm:

Marcos, the above was a reply to your second of three messages. Re the third message: you can define the notion of 'topological space' for countable sets as well as uncountable ones. I think it's often useful to do so, I bet someone here can tell us an application in Combinatorics or Algebraic Geometry. To distinguish N² and N you need some sort of extra structure - for example if you defined the operation + on N² and N (so that for N² (a,b)+(c,d)=(a+c,b+d) for example), then this isn't the same as defining + by using the + on N. In other words, if f:N² -> N is a bijection then defining (a,b)+(c,d)=f^-1 (f(a,b)+f(c,d)) will be different from the definition above.

Zacky Choo

Posted on Friday, 07 November, 2003 - 12:32 am:

Unfortunately not, there is no bijection from

.

You might think

and

are the same size, (i.e. they are infinite), so a bijection should exist. However, they have a different kind of infinity.

is countable,

is not.

The formal definition of countability means that the space must be a function of

. It matches the intuitive idea of counting.

To simplify matters, let us look at a stronger case of countability, called countedness The difference is, a set is counted if there exists a bijection to

For e.g.

is counted (and thus countable)
take the function f(z)=2z + 1 if z > = 0, f(z) = -2z if z < 0, where z in

. Check that it is a bijection

It takes a bit of thought, but a function can be found to biject

into

²

Now, an uncountable set is not countable. Surprise, surprise.

is uncountable, the proof is called "diagonalisation", but I'm too lazy now to write it out. And thus by definitions, no maps exist from it to

and thus no maps exist from

Ian Short

Posted on Friday, 07 November, 2003 - 10:19 am:

Zacky, Marcos acknowledges the uncountability of R, I'm sure. He is looking to see whether a bijection between R and R² is of any consequence; whether calculations in R² can be replaced by calculations in R given this bijection.

Markos- to make a general point. This identification of R and R² is ONLY a bijection. I say ONLY in that it need not necessarily preserve any properties of R and R² such as addition or topology, as David pointed out.

Unless this bijection, or isomorphism preserves the structure of R² with which you are dealing, it is of no use.

Ian

Marcos

Posted on Friday, 07 November, 2003 - 02:54 pm:

Thanks all...
Just two more questions:
Can anybody refer me to anything (an application, say) about what Dan was saying about the notion of a topological space when talking about countable sets?
On a related note, how do we show that we need at least n independant coordinates to describe n-space (in order to preserve structure)? Or is this in some way (that I'm missing) obvious?

Marcos

Kerwin Hui

Posted on Friday, 07 November, 2003 - 04:45 pm:

One nice example - consider the rationals Q with the usual (order) topology, i.e. sets of the form {x:a < x < b} are open and any union of such is open. This is not a discrete topology (since singletons are not open), and has some very cute properties. For example, consider the set Homeo(Q), i.e. continuous bijections Q to Q with continuous inverse. They form a group under composition, and this group is n-homogeneous for every n (i.e. any unordered n-subset can be mapped to any other unordered n-subsets), but it is not infinitely homogeneous, nor is it the full permutation group of Q.

The proof that R^m and Rⁿ , with their usual topology are different requires machinery that is usually in final year undergraduate level (i.e. (co)homology theory and/or homotopy theory). Basically, it shows that R^m -pt and Rⁿ -pt are different by considering how we can map the sphere of 1 dimension lower than min{m,n} to these punctured spaces we are interested in. For example, R² and R³ are different because if we perform the procedure of mapping a circle to R² -pt, some circles cannot be contracted to a point, but on R³ it is not very hard to see that we can always do that.

There are maps R² to R that will preserve the additive structure and some multiplicative structure (namely multiplication by rationals), but the proof requires the axiom of choice and so one cannot actually write down explicitly what the map is.

Kerwin

Marcos

Posted on Saturday, 08 November, 2003 - 09:50 am:

Thanks Kerwin,
(Now I'll go try to understand that first paragraph)

Marcos