Wikipedia:Reference desk/Archives/Mathematics/2011 February 1

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.

February 1

Can somone help with the following proof

Hi everybody/ I seem to have difficylties in understanding a specific thing in the followind proof of the following claime. Maybe somone can help? claim: There exist a countable non-weakly Frechet-Urysohn space. and here is the proof: Let x be an arbitrary point of the Stone-Cech reminder ω* of the discrete space ω. Then $X=\omega \cup \{x\}$ is the desired example.Indeed, it is a countable space. Let us assume that X is Weakly Frechet-Urysohn. As $x\in {\overline {\omega }}$ there must br a countable infinite disjoint familly F such that x(RZ)F. Let A and B be any two infinite subfamilies of F. Then both $x\in {\overline {\bigcup \mathbf {A} }}$ and $x\in {\overline {\bigcup \mathbf {B} }}$ must hold. But this is impossible if we choose A and B to be disjoint subfamilly of F as in this case $\bigcup \mathbf {A}$ and $\bigcup \mathbf {B}$ are disjoint subsets of ω and x is an ultrafilter of ω. The thing which I don't understand is this, Why does the fact that $\bigcup \mathbf {A}$ and $\bigcup \mathbf {B}$ are disjoint implies that it is not possible that both $x\in {\overline {\bigcup }}\mathbf {A}$ and $x\in {\overline {\bigcup }}\mathbf {B}$? Here are the required definitions: Definition: A point x is called weakly Frechet Urysohn point if whenever $x\in {\overline {A}}\setminus A$ there exists a countable infinite disjoint family F of finite subsets of A such that for every neighborhood V of x the subfamily $\{F\in \mathbf {F} :F\cap V=\emptyset \}$ is finite. If every point of a space is a weakly Frechet-Urysohn point then this space is called a Weakly Frechet Urysohn space. Definition: A point $x\in X$ and a countable infinite disjoint family F of X are said to be in the Reznichenko relation (Rz), written x(RZ)F, if the following holds: For every neighborhod V of x, the subfamily $\{F\in \mathbf {F} :F\cap V=\emptyset \}$ is finite. Thanks for any of you who will be able to help! Topologia clalit (talk) 17:27, 30 January 2011 (UTC)

For any set $S\subseteq \omega$ it holds that $x\in {\overline {S}}$ iff S is in the ultrafilter that represents x. This cannot be true for two disjoint sets. –Henning Makholm (talk) 14:58, 31 January 2011 (UTC)

Matching Socks in the Dark

You are in a dark room with no light. You have 19 grey socks and 25 black socks. What are the chances you will get a matching pair?

This is a question I came across on a list of wacky/difficult interview questions. Assuming it's not a trick (the question doesn't explicitly say you can only grab two socks, maybe I have a flashlight, etc.), I think the answer is 942/1122, but that seems too messy for a question like this. Anyone care to confirm/deny/correct?--68.51.73.79 (talk) 06:31, 1 February 2011 (UTC)

Doesn't seem messy. Calculate probability that both socks are grey. Calculate probability that both socks are black. Add together. Done. 71.141.88.54 (talk) 07:00, 1 February 2011 (UTC)

942/1122 is not too messy, but it's implausibly large. I got the same result at first -- it seems that we both thought that 19+25 was 34 . But it's really 44, so the chances are 942/1892. –Henning Makholm (talk) 09:38, 1 February 2011 (UTC)

[ec] I got ${\frac {19\cdot 18+25\cdot 24}{44\cdot 43}}={\frac {471}{946}}$. -- Meni Rosenfeld (talk) 09:39, 1 February 2011 (UTC)

Area of a quadrilateral

A Quadrilateral with sides taken in order in a plane ,Whose equations are given.What is the Formula for finding Area of Quadrilateral in terms of Coefficents of variables present in equaion of lines(sides).In fact i have find out a simple formula for area of triangle whose equation of sides are given .now I want to get a general formula.if no one has find out it ,then i will try for it for my interest. TRue Path Finder

First find the coordinates of the corners by solving each pair of equations for neighboring sides using Cramer's rule. Then insert into the shoelace formula. –Henning Makholm (talk) 19:49, 1 February 2011 (UTC)

I know this method ,but i want a general formula for quadrilateral.Also for Pentagon,hexagon ,... if equation of sides are given. — Preceding unsigned comment added by True path finder (talk • contribs) 02:32, 2 February 2011 (UTC)

Well, for a set number n of sides, substitute letters for the coefficients in your formulae, and follow Henning Malcolm's method. The result will be a formula for the area in terms of the equations' coefficients. (That doesn't work for an arbitrary number of sides. (Well, if you do it for a few values of n and see a pattern (but I doubt you will), then you can try to prove a more general rule, e.g. by induction on n. (How does the area change when we add a side?))) 99.40.234.78 (talk) 08:43, 2 February 2011 (UTC)