Mathematics desk
< August 3	<< Jul \| August \| Sep >>	August 5 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 4[edit]

The Stone Cech Compactification[edit]

Hi, Can ZANyone Explain why is the Stone-Cech Compactification, defined by the space of ultrafilters, is a compact space? —Preceding unsigned comment added by Topologia clalit (talk • contribs) 07:31, 4 August 2010 (UTC)[reply]

How are you making the Stone–Čech compactification out of ultrafilters? The only way I know only works for discrete spaces. Algebraist 09:23, 4 August 2010 (UTC)[reply]

The entry claims you can make it using maximal filters on the partial order of zero-sets. It's not hard to show that the original space embeds densely in this new space, that the new space is compact, and that it has the universal property. It's not clear to me, however, why it's hausdorff.203.97.79.114 (talk) 10:52, 4 August 2010 (UTC)[reply]

Suppose F and G are distinct maximal filters. Then there exists a zero-set A in F that is not in G. Since G is maximal, it must contain a zero-set B disjoint from A. Let f and g be functions with zero-sets A and B respectively. Then |f|−|g| is strictly negative on A and strictly positive on B. Setting U=(|f|−|g|)⁻¹(−∞,0) and V=(|f|−|g|)⁻¹(0,∞), we have that U and V are disjoint cozero sets containing A and B respectively, which correspond to disjoint basic open sets in the compactification containing F and G. Algebraist 11:56, 4 August 2010 (UTC)[reply]

Descrete spaces will be helpfull for me too.. Suppose we take the descerete case. I don't understand, Why is the space of ultrafilters defined on X is compact? In the topology we have mentioned a few days ago.. I think that in here it is specified that the space of ultrafilters defined over X is compact Hausdorff for any topological space X: http://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification Topologia clalit (talk) 09:57, 4 August 2010 (UTC)[reply]

The ultrafilter topology is generated by the basis of closed sets C_A={ultrafilters F such that A∈F}m where A ranges over subsets of X. Note that C_A∩C_B=C_A∩B, and similarly for any finite intersection. Suppose that we have an infinite family {A_i|i∈I} of subsets of X, such that any finite subset of the C_{A_i} has nonempty intersection. But then any finite subset of the A_i must likewise have nonempty intersection, so we can find a filter (and hence an ultrafilter) containing all of the A_i. This ultrafilter is in the intersection of the C_{A_i}, which is thus nonempty, and so the space of ultrafilters is compact. Algebraist 10:10, 4 August 2010 (UTC)[reply]

Thanks!!! Topologia clalit (talk) 10:09, 7 August 2010 (UTC)[reply]

{Ø}[edit]

What is {Ø}?199.126.224.156 (talk) 09:58, 4 August 2010 (UTC)[reply]

The set whose only element is the empty set. Algebraist 10:01, 4 August 2010 (UTC)[reply]

An alternative notation for which is {{}}. -- The Anome (talk) 13:52, 5 August 2010 (UTC)[reply]

Under a standard notation, it is also the number one, represented as 1.--Shahab (talk) 13:08, 4 August 2010 (UTC)[reply]

Geometry problem[edit]

Since the last programming competition question was answered so quickly, I think I'll try another one that has been bothering me for a long time. I asked before on the computing desk, but the answers didn't really answer the question. This is a paraphrased question from a programming competition I was in many years ago. Note, all numbers are decimal, not integer numbers.

Given three (x,y) coordinates, you have the definition of a triangle. Given three more positive non-zero numbers, you have defined three circles, each number being a radius. Is it possible for all three circles to fit in the triangle without overlapping the edges of the triangle or each other. Edges may touch, but not overlap.

My program was deemed correct, but I know it was wrong because I only tested the situation that the smallest circle goes into the smallest angle corner of the triangle and the largest circle goes into the largest angle corner - leaving the medium circle for the medium angle corner. There is another situation where the circles may be lined up along one side of the triangle, but not fit into the three corners. Is there a simple solution to this problem or is it required to be a complicated task of testing every possible layout for organizing the circles inside the triangle? -- kainaw ™ 12:19, 4 August 2010 (UTC)[reply]

There's something missing here. If the triangle is quite big and the 3 radii are quite small then they fit easily. If your phrase "Given three more positive non-zero numbers" is parsed "Given (three more) (positive non-zero numbers)" then that interpretation is possible. If it is parsed "Given three (more positive) non-zero numbers" then more positive than what - the other numbers are x,y pairs. What am I not seeing? -- SGBailey (talk) 20:56, 4 August 2010 (UTC)[reply]

Can't say I understand either. There's also no restriction about what planes the circles can be on, so the radii might stretch to the z-axis and allow a huge amount of circles. And what if you have cases where your circle has a huge radius? You can't ever fit that without overlapping something? Yea, I must be missing something too. --Wirbelwindヴィルヴェルヴィント (talk) 23:16, 4 August 2010 (UTC)[reply]

The question is to determine the values of the parameters for which it is possible to fit the circles inside the triangle. For example, if there were only one circle, the answer would be given by the radius r of the inscribed circle: if the radius of the circle is less than r, it can fit inside, otherwise not. For a pair of circles, one gets smaller triangles inside the big triangle for allowable locus for the center of each circle, and the solution is determined by calculating the maximal distance between a vertex of the first small triangle and a vertex of the second small triangle. Tkuvho (talk) 23:41, 4 August 2010 (UTC)[reply]

The question uses "given" to mean "you are given". So, you are given something like: (1,4.5) (-3.2,9.3) (0,45) 23 41.6 21. The first three x,y coordinates define a triangle. The second three numbers define three circles. Will those three circles fit in the triangle? It isn't a question about one specific case. It is a question about coming up with an algorithm that will work for EVERY possible set of x,y coordinates and radius values. -- kainaw ™ 00:01, 5 August 2010 (UTC)[reply]

You wrote: "There is another situation where the circles may be lined up along one side of the triangle, but not fit into the three corners." Can you provide an example? -- 119.31.126.81 (talk) 12:37, 5 August 2010 (UTC)[reply]

Triangle sides 1, 100, 100; Circle radii 0.9, 0.9, 0.9. -- Meni Rosenfeld (talk) 15:38, 5 August 2010 (UTC)[reply]

Exactly. So, the only arrangements that I can think of are placing the circles into the three corners and checking for overlap or placing the circles in a line and checking for overlap. I don't know of any other arrangement for the circles that would work when those two fail. I also can't think of a shortcut to make this an easier problem. (Also, the original problem was even more complicated. They weren't just circles. There was two parallel planes. The triangle extended over both planes to make a triangular container. The circles had a separate radius on each plane, making them look like little hats with one wide radius and one smaller radius. You were allowed to flip the circles over to try and make them fit better. So, I figure the real problem was performing the "does it fit" algorithm and flipping the circles if they don't fit. That is why I assume that the "does it fit" algorithm should be easy.) -- kainaw ™ 15:53, 5 August 2010 (UTC)[reply]

Sloppy Notation?[edit]

I'm presently reading through a set of notes on Vector Calculus and have met a result that I understand but I'm not sure that the notational rules have been observed.

$[\nabla \times {\frac {\mathbf {x} }{r^{3}}}]_{i}=\varepsilon _{ijk}{\frac {\partial }{\partial x_{j}}}{\frac {x_{k}}{r^{3}}}=\varepsilon _{ijk}{\frac {\delta _{jk}}{r^{3}}}-3\varepsilon _{ijk}{\frac {x_{k}}{r^{4}}}{\frac {x_{j}}{r}}=0$

In the term with both an epsilon and a delta, I was under the impression that you're not allowed to let j=k and then sum over j in the delta and simultaneously leave the epsilon, by which the delta is multiplied, unchanged. Is this correct notation? Thanks asyndeton talk 13:40, 4 August 2010 (UTC)[reply]

Looks like a typo. Final expression Should be

\varepsilon _{ijk}{\frac {\delta _{jk}}{r^{3}}}-3\varepsilon _{ijk}{\frac {x_{k}}{r^{4}}}{\frac {x_{j}}{r}}=0

Gandalf61 (talk) 14:43, 4 August 2010 (UTC)[reply]

My mistake, I just copied and pasted

\delta _{ij}

without amending it for my purposes. But my original question still stands, unless you're implying that having

\delta _{ij}

was the only mistake in the whole thing? asyndeton talk 15:22, 4 August 2010 (UTC)[reply]

I can't make any sense of your question. Where has anyone let j=k and then summed over j in the delta and simultaneously left the epsilon, by which the delta is multiplied, unchanged? Algebraist 15:46, 4 August 2010 (UTC)[reply]

(edit conflict) Yes, with

\delta _{jk}

instead of

\delta _{ij}

, the expression is correct. The step that is omitted is:

\varepsilon _{ijk}{\frac {\partial }{\partial x_{j}}}{\frac {x_{k}}{r^{3}}}=\varepsilon _{ijk}\left({\frac {1}{r^{3}}}{\frac {\partial x_{k}}{\partial x_{j}}}+x_{k}{\frac {\partial }{\partial x_{j}}}{\frac {1}{r^{3}}}\right)

Of course,

\varepsilon _{ijk}\delta _{jk}

is 0 for all i, since

\varepsilon _{ijk}

is 0 if j = k and

\delta _{jk}

is 0 if j ≠ k. But this is just saying that

\nabla \times \mathbf {x}

is the zero vector everywhere~~, which is intuitively clear because, considered as a vector field, $\mathbf {x}$ is purely radial and so has no rotation~~. Gandalf61 (talk) 15:56, 4 August 2010 (UTC)[reply]

Homeomorphisms of R³[edit]

I was thinking about quotient spaces of the sphere. Take the sphere an identify the North and South poles. This can be done many ways; but I have two examples in mind. We can take the poles and pull them away from the centre, pull them around outside of the sphere and then glue them together. This will give a pinched torus (see the left hand image). The other way would be to push the poles into the sphere towards the centre, and then glue them when they meet (see the right hand image). I can see that these are embeddings of the quotient space S²/S⁰ into R³. I was wondering:

Is there a homeomorphism of R³ that takes one to the other?
Given two embeddings of ƒ, g : X → Y, can we find a homeomorphism h : Y → Y that carried ƒ(X) to g(X)?

— Fly by Night (talk) 16:08, 4 August 2010 (UTC)[reply]

The figure on the left has the property that the (open) interior of the pinched torus is homeomorphic to a 3-ball. The figure on the right has the property that the interior of the torus is homeomorphic to a circle times a disk. Therefore there is no such homeomorphism. Tkuvho (talk) 17:15, 4 August 2010 (UTC)[reply]

The image on the right isn't a torus. There's no hole in the middle. Denote it by A then the fundamental group is π₁(A) ≅ Z. The loops around the outside of the outside of A can be deformed into a point. (The inner circle of a torus has been identified, i.e. is just a single point. The singularity of the image.) I think that the interior of the object on the right is homeomorphic to the interior of a solid torus. So you would be right: there is no homeomorphism. Thanks a lot for your comments. — Fly by Night (talk) 18:01, 4 August 2010 (UTC)[reply]

This raises another question. Denote the pinched torus by P and the object on the right by A. It seems that the interior of P is homeomorphic to the exterior of A and the interior of A is homeomorphic to the exterior of P. Is there a name for this property? Have these kind of objects been studied before? — Fly by Night (talk) 18:09, 4 August 2010 (UTC)[reply]

If you complete R³ to S³ by adding a point at infinity, then the two figures are identical, modulo switching interior and exterior. You might be interested in related material at surgery theory. Tkuvho (talk) 19:26, 4 August 2010 (UTC)[reply]

But by adding a point at infinity we change the topology of the host space. Besides, it's special to R³. Have you any more general thoughts? Say that a space X is embedded in a space Y, then how many possible, non-homeomorphic images are there? Given two embeddings ƒ, g : X → Y of a compact 2-dimensional space X into a 3-dimensional space Y then what does in mean for the interior of ƒ(X) to be homeomorphic to the exterior of g(X) while the interior of g(X) is homeomorphic to the exterior of ƒ(X)? — Fly by Night (talk) 19:36, 4 August 2010 (UTC)[reply]

Adding a point at infinity is not special to R^3: see One-point compactification.—msh210℠ 19:43, 4 August 2010 (UTC)[reply]

Okay, fine. Could you now please address the question? — Fly by Night (talk) 21:13, 4 August 2010 (UTC)[reply]

Note that the exterior of the figure on the right is homeomorphic to a punctured ball, i.e. a 2-sphere times an open interval. Tkuvho (talk) 08:38, 5 August 2010 (UTC)[reply]

A 2-sphere times an open interval is an open thick shell. I've never heard of that described as a "punctured ball"; it seems an odd name. You are correct in your description of the exterior, though. It could also be described as R³ minus an open ball. --Tango (talk) 14:25, 5 August 2010 (UTC)[reply]