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November 4

Lindelöf Spaces and Second Countable Spaces

So, I have studied both of the articles Lindelöf space and Second-countable space. I know that a space X is Lindelöf if every open cover has a countable subcover and that X is a second countable space of its topology has a countable basis. I also know that the a second countable space is Lindelöf but in order for a Lindelöf space to be second countable, our space in question must be a metric space.

My question is, why is this restricted to metric spaces? It seems to me that I can prove Lindelöf and second countable are equivalent for any topological space and here is the proof. I will only do it one way because the other way we already know is true for general topological spaces.

Let X be a Lindelöf space. This means that ever open cover has a countable subcover which means that in particular the topology on X has a countable subcover and let that countable subcover be the basis and we are done.

I know that there is a counterexample listed here too in the articles but I don't understand why it is a counterexample and can anyone please explain what is wrong in my reasoning? It looks to me that the two conditions are equivalent no matter what the space is.
A Real Kaiser 00:21, 4 November 2007 (UTC)[reply]

Seems like you're confused about what a base is. The topology on X certainly does have a countable subcover: in fact it has a finite subcover, namely the singleton set of X itself. However, unless X has the trivial topology, there are other open sets besides the empty set and X itself, so {X} is not a base of X. A set of subsets of X is only a base if every open set of X can be expressed as a union of those subsets. —Keenan Pepper 00:49, 4 November 2007 (UTC)[reply]

You are absolutely right. It just skipped my mind that I am not looking for a mere subset but rather a basis. Okay, so now the question is how can one prove that if a metric space X is Lindelöf then it is second countable?
A Real Kaiser 05:11, 4 November 2007 (UTC)[reply]

Apply the Lindelof property to the set of all balls of size 1/n . Then take the union of the sets you got as n goes through the natural numbers. You get a countable union of countable sets, so it's countable. This will be a base for the space. 79.182.230.190 18:20, 4 November 2007 (UTC)[reply]

Bernoulli numbers

In the first paragraph of the Bernoulli number page it is written that there is no known elementary description for Bernoulli numbers .What is the meaning of "elementary description"? I thought that it meant that there is no mathematical rule to find the nth Bernoulli number that is not a sum of an infinite series because there was none on the page .but then I found somewhere on the internet a simple definition that is not a sum of an infinite series and also it does not depend on previous values of $B_{n}$ .So, my question is :what does "elementary description" mean ? --George 04:04, 4 November 2007 (UTC)[reply]

I suppose the meaning intended here is that there is no definition in the form of an expression using only elementary functions, in other words, a closed-form solution of a defining equation for the Bernoulli numbers. --Lambiam 06:38, 4 November 2007 (UTC)[reply]

George - did you say you have somewhere seen a finite and non-recursive expression for

B_{n}

? Do you remember where that was ? The simplest "rule" for finding the value of

B_{n}

that I can think of is

B_{0}=1

B_{n}=\left({\frac {-1}{n+1}}\right)\sum _{j=0}^{n-1}{n+1 \choose {j}}B_{j}{\mbox{ ; }}n>0

but that is recursive, so it probably doesn't qualify as an "elementary description". Gandalf61 09:57, 4 November 2007 (UTC)[reply]

Might it have been a generating function you saw? They're easy to mistake for formulas. Black Carrot 06:42, 5 November 2007 (UTC)[reply]

The formula is

B_{n}=\sum _{k=0}^{n}\sum _{j=0}^{k}j^{n}(-1)^{j}{\frac {\binom {j}{k}}{k+1}}

I'm not sure if this can be considered an elementary description ,if it's not one please explain the reason .Thanks --George 10:20, 5 November 2007 (UTC)[reply]

Since j ranges from 0 to k, should

{\binom {j}{k}}

in the numerator actually be

{\binom {k}{j}}

instead ? Gandalf61 10:55, 5 November 2007 (UTC)[reply]

Well, the Gamma function is non-elementary, so in some sense the binomial coefficient function is not elementary, either. This might have some relevance to the claim that no elementary formula exists. -- Meni Rosenfeld (talk) 13:11, 5 November 2007 (UTC)[reply]

yes, I'm sorry, it is

B_{n}=\sum _{k=0}^{n}\sum _{j=0}^{k}j^{n}(-1)^{j}{\frac {\binom {k}{j}}{k+1}}

, I just mistyped the formula. And I think that the binomial cofficient function is elementary because it is used here for integers only, and it can be calculated in a finite number of multiplications without using the gamma function. --George 14:59, 5 November 2007 (UTC)[reply]

Apparently, whoever claimed the non-existence of an "elementary description" was mistaken. I've removed that claim. Could you add the explicit formula to the article? I don't know the source. --Lambiam 10:50, 7 November 2007 (UTC)[reply]

This is just a different way to write the standard recurrence formula. I would not say that rewriting it in this form provides an "elementary description". It is not "elementary" in the sense of containing only algebraic/exp-log functions, and series are not usually considered "closed-form" (double series much less so). The formula is listed on MathWorld, by the way. Fredrik Johansson 12:24, 7 November 2007 (UTC)[reply]

I can't remember the source but you can use the link to mathworld. Thanks to everyone who answered this question , it puzzled me for a very long time.--George 20:21, 7 November 2007 (UTC)[reply]

Terminology: Points in the plane

I'm working on something that slightly exceeds my grasp of mathematics, and I need some hints on terminology, or a family of mathematics that would be helpful in dealing with this.

Say you have n points in the plane (in the order of a thousand), and no pair has a distance smaller than d. My intuition is that if you want to place these points to occupy the minimum surface area, you need to place them at the vertices of equilateral triangles, sides of length d, that tessellate the plane. It sounds like something would have been proved a long time ago, but I can't find it anywhere. Anybody know a name for this, or something like this? risk 04:19, 4 November 2007 (UTC)[reply]

Howdy, I think you are looking for sphere packing, except it might be better called circle packing. Since the distance between any two points is at least d, that means each point is surrounded by a circle of diameter d, and none of the circles properly overlap, though they can be tangent. I don't know a thing about sphere packings, but I think in two dimensions your description is about right. There is even a picture in the article showing the densest packing. JackSchmidt 04:42, 4 November 2007 (UTC)[reply]

Yes, the densest packing in the plane is with circles\hexagons\equliateral triangles, however you wish to call them. However, if you want a meaningful answer for the case of finitely many points, you will have to define "minimum surface area". Perhaps you mean the area of the convex hull of the points? -- Meni Rosenfeld (talk) 13:17, 4 November 2007 (UTC)[reply]

The arrangement of 1000 points on a plain with the smallest possible convex hull is along a straight line, giving zero area. —Preceding unsigned comment added by 84.187.90.99 (talk) 23:06, 4 November 2007 (UTC)[reply]

As you are essentially looking for an optimal circle packing, there is some inner logic in looking for the arrangement with minimum circumscribing circle. —Preceding unsigned comment added by 84.187.90.99 (talk) 23:18, 4 November 2007 (UTC)[reply]

Thanks, these answers helped. The notions of density and circle packing were what I was looking for. I thought that I was thinking of the minimum convex hull, but as .99 pointed out, that makes a straight line optimal, which isn't what I was looking for. Perhaps something along the lines of minimizing the average distance between two points. Anyway, now that I have what I need, the shape of the points together isn't that relevant anymore. risk 00:43, 5 November 2007 (UTC)[reply]

Simple Maths

What is 1+1?

If you could answer this question I will be forever greatful!!! Because it is way too hard to figure out. —Preceding unsigned comment added by 220.233.83.26 (talk) 05:35, 4 November 2007 (UTC)[reply]

Did you read our 1+1 article? It doesn't seem that hard to me. As a bonus, see Wikipedia:Reference desk/Archives/Mathematics/2007 April 18#Why do 1+1=2? and Wikipedia:Reference desk/Archives/Mathematics/2007 April 30#1+1=2 ... why???. --Lambiam 06:45, 4 November 2007 (UTC)[reply]

A television channel? Titoxd^{(?!? - cool stuff)} 20:10, 4 November 2007 (UTC)[reply]

Depends on your definition of multiplication, but if you are in a ring or field, you can set 1 + 1 = a, and note necessarily that a * a = a + a (if "1" is the multiplicative identity). SamuelRiv 14:22, 4 November 2007 (UTC)[reply]

We can choose the answer to be whatever we like, so long as we are prepared to live with the consequences. A friend who attended a Roman Catholic girls' school many years ago claimed you could never go wrong with the explanation "Because God in his infinite wisdom made it that way." I doubt that would have sufficed in a Jesuit boys' school. I call to your attention the following fact: Research has confirmed that a number of animals, some not even great apes or primates, are capable of doing this calculation. What that says about your abilities, I am not qualified to say; I would suggest that you seek professional counseling. --KSmrq^T 16:23, 4 November 2007 (UTC)[reply]

I would suggest that YOU seek professional counseling. Answer the question.

ok. The Context decides the answer.

In Arithmatic. 1+1=2. i.e. the combination of unary operators becomes a pair.
In the Binary Number system, 1+1=10b2
In the Uninary Number system, 1+1=11 ( see just like the Roman numerals ).
In Set theory, 1 is an identity element, so 1+1=2(1)s
In Algebra, the communitive property of addition says 1+1=1+1
In Finite Math, The sum of two infinities is infinity. 1+1=1
Of course you cannot dismiss philosophy. In Philosophy, the idea of 1+1, is that 1 is a unique idea or property, and 1+1 is that it is no longer unique, or that it is in summation with its mirror image. ( look at an apple in the mirror. You see two apples, 1+1, but how many Apples are there? 1, so 1+1=1, i.e. 1 view + 1 view = 1 view ( larger ).

- And lastly, In the early part of the 1900s, several mathamatitions ( namely David Hilbert, the infinate spaces guy ) worked on something known as Formalism. A very deep and structured way of looking at exactly what we mean by 1+1. It ususally took several hundered pages of ideas just to get to 1+1, but Kurt Godel came along, and proved that in any system of formalization, it was impossible to prove any arbitrary theorm, so 1+1=? ( go slow on this one )

"Sufficiently expressive consistent axiom systems can never prove their own consistency"

[Philosoply Of Mathematics:Formalism]

Artoftransformation 13:33, 9 November 2007 (UTC)[reply]

In Boolean algebra, if you're using '+' for inclusive disjunction, 1+1=1; if you're using '+' for exclusive disjunction, 1+1=0. (I've seen both conventions.) —Tamfang 04:20, 10 November 2007 (UTC)[reply]