Difference between R^2 and R: What distinguishes them in terms of topology?

[pervect]

What makes R^2 different from R? Unless I'm mistaken (which is unfortunately quite possible), R^2 has the "same number" of points as R.

I'm thinking that what makes R^2 different from R is the topology, which is why I picked this forum. But what sort of conditions do we really need to make R^2 distinguishable from R? Having more of a physics background than a math background, I think of imposing a metric on the space, though I'm not sure if this is overkill. Is there a less stringent notion that we can use to distinguish R^2 from R?

I'm also a bit confused over what's needed to impose a metric. Under metric topologies, the only requirement listed on mathworld,

http://mathworld.wolfram.com/MetricTopology.html

is that the space be Hasdorff. But there are also some entries about T1 spaces and T2 spaces. I'm not clear about what the difference between them is supposed to be, but it mentions that T1 spaces are supposed to be metrizable, and nothing is said one way or the other about the metrizability of T2 spaces

http://mathworld.wolfram.com/T1-Space.html http://mathworld.wolfram.com/T2-Space.html

 

[Lonewolf]

To quickly address your last remark, if a topological space satisfies the T2 axiom, then it also satisfies the T1 and T0 axioms. i.e. T2 implies T1, and T1 implies T0.

 

[matt grime]

R and R^2 are non-isomorphic vector spaces

they are isomorphic sets (they have "the same number of points")

there is no homeomorphism between R and R^2 with the usual metric topologies, one flavour of de rham cohomology distinguishes between them (compactly supported?)

two real finite dimensionaly vector spaces are homeomorphic iff they have the same dimension, though I can't think of an elementary proof in general.

in this case, imagine a point removed from R, 0 wlog, then if there were a homeomorphism this would pass to a homeomorphism to between R\0 and, say R^2\(0,0) (assume the origin maps to the origin wlog), but the first set is disconnected, the second isn't.

 

[chronon]

I'm not an expert at this, but I did find the following

http://mathworld.wolfram.com/LebesgueCoveringDimension.html

and

http://en.wikipedia.org/wiki/Lebesgue_covering_dimension

 

[HallsofIvy]

As far as "imposing a metric" is concerned, you need to look at
"metrizable topology":
http://mathworld.wolfram.com/MetrizableTopology.html

 

[gravenewworld]

Dimension in mathematics means how many vectors there are in the basis of a vector space. So a simple answer is that R has only 1 vector in its basis while R^2 has 2 vectors.

 

[pervect]

OK, thanks to all who responded. I'm going to have to take some time to digest this fully, but I do have a few remarks. Even more as I start to write...

I do think that the vector space explanation is the simplest, but I think it's one of the least powerful. The reason that I think it's less powerful is that one has to know how to add two vectors together to define a vector space. That's a fair amount of extra "structure".

It seems to me that "neighborhood" is a more primitive concept than vector addition. Without an idea of neighborhood, I don't think the notion of continuous transformations can really be defined. So as far as the homeomorphisms go, I don't see how we distinguish an arbitrary mapping between R^2 and R that's discontinuous and thus not a homeomorphism until we define the idea of neighborhood.

I didn't quite grasp the answer about cohomologies, alas. I've heard the term used, but the definitions didn't click then, and re-reading the appropriate web pages on Mathworld didn't seem to help much :-(.

Thanks for the clarification about how T2 implies T1 implies T0. That was very helpful.

The "Lebesque covering dimension" is especially interesting, I'm not sure I fully grasp it yet, though, but it looks like it might be the answer I was looking for.

Again, thanks to all who responded.

 

[Hurkyl]

The "simple" proof that R and R^2 are not homeomorphic is the chicken's foot. Take a subset of R^2 in the shape of a Y, and map it continuously to R. No matter how you do it, at least two of the lines are going to overlap.


You're absolutely right that you can't talk about continuous functions without neighborhoods: the very definition of continuous is how it behaves with respect to open sets.

I would disagree that neighborhoods are more fundamental than vector addition: they're simply different.

 

[matt grime]

Are you considering R^2 as merely a set or a vector space. If it isn't a vector space then you should say what dimension you think you are talking about.

 

[HallsofIvy]

Dimension in mathematics means how many vectors there are in the basis of a vector space. So a simple answer is that R has only 1 vector in its basis while R^2 has 2 vectors.

That's dimension of a vector space. You can define the dimension of a set or topological space in a number of different ways.

 

[HallsofIvy]

By the way, if I remember correctly:
T0: every singleton set is closed.
T1: given any two points there exist an open set containing one but not the other.
T2: given any two points there exist 2 disjoint open sets so that each contains one of given points.

It is obvious that T2 implies T1 and fairly easy to prove that T1 implies T0.

The "T" is from Tychinov, a Russian mathematician, and "T2" specifically is also called "Hausdorff" after a German mathematician.

 

[S. Carton]

Are you considering R^2 as merely a set or a vector space. If it isn't a vector space then you should say what dimension you think you are talking about.

Isn't this a problem of terminology more than anything? IIRC, R^2 is shorthand for Euclidean 2-space; the Euclidean metric is therefore implicit, and more or less the baseline for what 2-dimensional actually means.

You can compare spaces to it with the topology induced by the metric, which is defined in R^2 BY DEFINITION OF R^2 already.

IOW, if you consider R^2 as "merely a set" in which you have to "say what dimension you think you are talking about" then it ain't R^2 anymore, is it?

 

[pervect]

Are you considering R^2 as merely a set or a vector space. If it isn't a vector space then you should say what dimension you think you are talking about.

I wasn't sure what I was looking for when I set out, but I came to realize that what I was looking for was a concept of dimension that applied to a topological space.

It looks like the Lebesgue covering dimension fits the bill. Unfortunately, I still don't fully understand, for example, why one needs exactly 3 refinements to cover a 2-d open set.
http://www.campusprogram.com/reference/en/wikipedia/d/di/dimension.html

 

[Hurkyl]

It's not 3 refinements, it's an order 3 refienment. (Was order the right word?)

I haven't wrapped my mind fully around it, but here's an example of sorts:

Take the 2-D plane: you can tile this with regular hexagons. You can make this into an open covering by taking as your open sets a neighborhood of the hexagon (I.E. the union of all the e-balls centered at a point in the hexagon).

The order of this covering is 3; points near the vertex of a hexagon will lie in three open sets, points near an edge willl ie in two, and the rest lie in only one set.

 

[mathwonk]

resources: "dimension theory" by hurewicz and wallman, "why space has 3 dimensions" by poincare.

poincare's essay is for the general public on the notion of dimension. he says basically that he calls a finite set zero dimensional for starters. then a set is 1 dimensional if it can be separated by removing a zero dimensional set. e.g. as matt grime pointed out, R^1 is disconnected by the removal of anyone point, hence is one dimensional.

R^2 is not disconnected by removing one point, but is disconnected by removing a copy of R^1 hence R^2 is two dimensional. etc etc..

It has been 40 years since i read it, but i believe hurewicz and wallman defined a topological space to be n dimensional at a point, if there is a basis? of neighborhoods of that point with n-1 dimensional boundaries? so R^1 has dimension one at the origin since there is a local neighborhood basis of intervals, each with boundaries which are just 2 points.

I also like the lebesgue covering dimension definition. It is especially surprizing to me that it is true, and interesting that anyone ever thought of it. it is also useful for defining a special cohomology that is used in algebraic geometry.

ordinary cohomological dimension is based on things being boundaries of other things, except one needs some further conditions, like compactness, since e.g. all R^n spaces for all n, have trivial ordinary cohomology.

The fact that a sphere has dimension 2, is based in this case on the fact that all loops in a sphere are boundaries of discs, but there is a surface, namely the sphere itself, which does not bound anything, within the sphere itself. here one essentially takes some standard objects as representing each dimension, like families of line segments are dimension one, families of triangles are dimension 2, families of tetrahedra are dimension 3, etc etc, and then one maps these standard objects into other spaces.

The lebesgue covering dimension comes into play in the definition of so called cech cohomology, with various coefficients, ultimately "sheaves". Open sets are used to imitate polyhedral structure as follows: A "vertex" is just an open set, an "edge" is a pair of open sets with non empty intersection. A 2 dimensional face is three open sets with non empty intersection. An n dimensional face is a collection of n+1 open sets with non empty interscetion.

Then lebesgue's theorem says that on an n dimensional space a suitably refined open cover will represent an n dimensional polyhedron, since more than n+1 sets will never have an non empty intersection.


Note for example that the structure of an ordinary n dimensional polyhedron is faithfully represented this way by the open cover consisting of the open "stars" of the vertices.

e.g. an interval is recaptured from the two open sets which are the complements of the two end points. a triangle is captured by the open cover via three open sets which are the complements of the three sides. i.e. each such open set represents the one vertex it contains, and the intersection of two of these represents the one (open) edge it contains, and the intersection of all three represents the face of the triangle.

cech cohomology is defined by assigning coefficients to each such cech polyhedral face, and making some algebraic constructions.
cech sheaf cohomology is defined by using coefficients which assign to each cech "face" the family of functions defined on that open set and having some property, like holomorphicity, or rational functions, etc...

one can also use more general "functions" as coefficients such as sections of a line bundle or vector bundle. In all cases the lebesgue dimension of the space gives the dimension above which the cohomology groups can not be non zero.

Another definition of dimension is based on filtrations by subspaces. in a vector space the dimension is the length of a maximal nested family of subspaces. I.e. point, line containing the point, plane containing the line, etc,... If one gets n+1 such ensted subspaces at most, then the dimension is n.

This generalizes in algebraioc geometry to the definition of the dimension of a noetherian space. I.e. the dimension of an algebraic variety is the length of a maximal nested chain of algebraic subvarieties. pont, curve containing the point, surface containiong the curve, etc etc...


I think a noetherian space is one in which descending chains of closed sets are always finite. oh yesy that's right because in affine algebraic geometry that correspond to ascending chains of ideals in the coordinate ring being finite, the usual meaning of noetherian.

thus the corresonding definition of dimension in a ring, is the maximal length of a nested chain of prime ideals, since these correspond to irreducible sets. oops that emans i forgot to say irreducible in the definition above.


for rings this is called the "krull" dimension.

anyway dimension is a big subject, and a central one in all geometry.

 

[mathwonk]

linear dimension also has another wrinkle in nifinite dimensions, e.g. in hilbert space where there is not only a linear vector satructure but a scalar product. then since we can measure angles there are two notions of independence, orthogonality and usual linear independence. then we copuld define the dimension a=s the cardinality of amximal lienarly independent set, or as the cardinaloity of a mximal orthonormal set, and that can be smaller. i.e. the most classical hilbert space "little L2", would have uncountable dimension in the first sense but countable dimension in the second.