Properties of Subsets of a Sphere - Closeness and Boundness

[pivoxa15]

I need to show properties of subset of a sphere.

Would it be okay to project each point in the sphere onto the real number (by converting them to a magnitude - like vector magnitude) and deduce properties of the subset of the points on R and imply them back to the subset on the sphere? Would they be equivalent? Is it a legimate thing to do?

The properties are closeness and boundness.

The reason is it is much easier to work with R (real number line) than a 3D sphere.

 

[arildno]

Consider the polar coordinate representation of the sphere: r=R.
Remember that the angles vary between: [itex]0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi[/tex]

Now, ask yourself:

1. What sort of figure is the sphere as represented in $(r,\theta,\phi)$-space?

2. Is the transformation of this figure into a sphere in (x,y,z)-space a continuous transformation, and if it is, how does that relate to the properties of closedness and boundedness?

 

[benorin]

It is not of neccessity that a subset of a sphere be closed.
For example: the subset consisting of the sphere less a single point.
Furthermore, there exists a bijection between the set of points on a sphere and the set of points of an infinite plane, and hence betwixt any two subsets thereof. Use, for example: stereographic projection.

 

[pivoxa15]

Thanks for the suggestions but I don't want to complicate things too much. What about my original proposition? Is it valid?

 

[HallsofIvy]

What do you mean by "project them on real number". A lot depends on exactly how you do it. Also, it is not clear to me what you mean by "on a sphere". On the surface of a sphere or including the interior? If you have to be careful that, in your projection, you don't lose any information. For example, if your set is on the surface of a sphere centered at the origin of a coordinate system, and you replace each point (x,y,z) by its norm, then every point is "projected" to the same number!

 

[arildno]

It is not of neccessity that a subset of a sphere be closed.

I never said it was, but your clarification was in order. Thanks.

 

[pivoxa15]

What do you mean by "project them on real number". A lot depends on exactly how you do it. Also, it is not clear to me what you mean by "on a sphere". On the surface of a sphere or including the interior? If you have to be careful that, in your projection, you don't lose any information. For example, if your set is on the surface of a sphere centered at the origin of a coordinate system, and you replace each point (x,y,z) by its norm, then every point is "projected" to the same number!

I meant the exterior of the sphere. So you think I lose information when I project every point to the real numbers through the Norm? The norm is calculated from all (x,y,z) for each point however.

 

[HallsofIvy]

Now that makes even less sense! Do you mean the SURFACE of the sphere? (Which is NOT the exterior!). And what do you mean by "the norm is calculated from all (x,y,z) for each point however"? What do you mean by "norm from (x,y,z) for each point". Do you mean the length of the vector from some (x,y,z) to the point on the surface of the sphere?
And if you really calculate that for every "all (x,y,z)" for each point, then you will have the set of all real numbers corresponded to every point. That makes no sense to me.

 

[pivoxa15]

Now that makes even less sense! Do you mean the SURFACE of the sphere? (Which is NOT the exterior!). And what do you mean by "the norm is calculated from all (x,y,z) for each point however"? What do you mean by "norm from (x,y,z) for each point". Do you mean the length of the vector from some (x,y,z) to the point on the surface of the sphere?
And if you really calculate that for every "all (x,y,z)" for each point, then you will have the set of all real numbers corresponded to every point. That makes no sense to me.

I meant the surface of the sphere. The exterior of the sphere would mean the outside of the sphere wouldn't it?

For the norm I meant the distance of each point on the surface to the origin. I admit that the set of these distances will not preserve any information about the kind of subset on the sphere. But as a result of doing the mapping, is it enough to show that the set of points on the sphere is closed and bounded because the set of the norm of points is closed and bounded?

 

[benorin]

FYI: a set of points is bounded if and only if there exists a sphere which contains them, so that is trivial.

 

[Euclid]

An easy way (in this case) to show it is closed is to show its complement is open. (I assume we've agreed you're talking about the boundary of a solid sphere.)

 

[benorin]

Stereographic projection

Let us construct a mapping taking points on a sphere to points in a plane (an infinite plane):

Let the sphere be of radius R, (and, for ease of construction, let it be in a coordinate system, say R^3,) having equation

$$x^2+y^2+z^2=R^2$$​

The set of points on the sphere is the domain of our mapping; the range will be the xy-plane. Let the point N(0,0,R) denote the North pole of the sphere. Map points as follows: for every point P(x,y,z) on the sphere, construct the ray NP and let the point where the ray NP intersects the xy-plane be P'(x',y',0): this is the mapping, namely P->P'.

This mapping is called stereographic projection, it is a bijection (if you adopt the "point at infinity" convention.) Points on the upper hemisphere get mapped outside of the circle of radius R, and points on the lower hemisphere get mapped inside of the circle of radius R; also, the point N(0,0,R) gets mapped to the so-called "point at infinity."

Now all the properties of subsets of points on a sphere are those of subsets of points in the plane. Nice