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September 19

Cohen and the axiom of choice

The article on Paul Cohen mentions that:

Cohen is noted for developing a mathematical technique called forcing, which he used to prove that neither the continuum hypothesis (CH), nor the axiom of choice, can be proved from the standard Zermelo–Fraenkel axioms (ZF) of set theory.

I am trying to find out the exact year in which Coehn first proved the result concerning the axiom of choice (not the continuum hypothesis) and the year and the paper when it was first published, if ever. Thanks--Shahab (talk) 11:21, 19 September 2012 (UTC)[reply]

In 1963 according to this:

In 1942 Gödel attempted to prove that the axiom of choice and continuum hypothesis are independent of (not implied by) the axioms of set theory. He did not succeed, and the problem remained open until 1963. (In that year, Paul Cohen proved that the axiom of choice is independent of the axioms of set theory, and that the continuum hypothesis is independent of both.)

It was published in two parts, but the result had previously appeared in april 1963 and were presented may 3, 1963 at Princeton, according to the booknote at end of second paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC300611/?page=6

First paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC221287/?page=1

To work out where exactly the independence of ZFC of the axiom of choice is proven, I would first have to figure out what all this means...

Other info: http://www-history.mcs.st-andrews.ac.uk/Biographies/Cohen.html Ssscienccce (talk) 17:26, 19 September 2012 (UTC)[reply]

Solve this please

(Moved by User:JackofOz from Miscellaneous Ref Desk)

Questions 1 - 5

There are six soccer teams - J, K, L, M, N and O - in the Regional Soccer League. All six teams play each Saturday at 10 a.m. during the season. Each team must play each o f the other teams once and only once during the season.

Team J plays team M first and team O second.
Team K plays team N first and team L third.
Team L plays team O first.

On the first Saturday, which of the following pairs of teams play each other ?

(A)    J and K ; L and O ; M and N
(B)    J and K ; L and N ; M and O
(C)    J and L ; K and N ; M and O
(D)    J and M ; K and L ; N and O
(E)    J and M ; K and N ; L and O

Which of the following teams must K play second ?

(A)    J
(B)    L
(C)    M
(D)    N
(E)    O

What is the total number of games that each team must play during the season ?

(A)    3
(B)    4
(C)    5
(D)    6
(E)    7

— Preceding unsigned comment added by 175.110.112.185 (talk) 19:43, 19 September 2012 (UTC)[reply]

You are essentially told the answer to the first question directly. I'm tempted to say it's obviously (E) (can you see why that's the case?)

Question two is done by process of elimination. K does not play N or L second because it plays them first and third respectively. J and O compete against each other in the second match so neither of them can play K. So the answer is M.

The answer to the third question is 5, obviously. For each other team, there are 5 other teams, and each team must compete against each other team exactly once. Widener (talk) 20:22, 19 September 2012 (UTC)[reply]

Creating a chart typically helps on this type of problem:

        O P P O N E N T S
TEAM   1st 2nd 3rd 4th 5th
----   --- --- --- --- ---
 J
 K
 L
 M
 N
 O

Try filling that in with what you know already. (Although, this doesn't look like so much of a logic problem as a reading comprehension problem.) StuRat (talk) 00:37, 20 September 2012 (UTC)[reply]

Show that a series is not Absolutely convergent.

Show that this series is not absolutely convergent: $\sum _{n\neq 0}e^{inx}{\frac {1}{2\pi in}}(e^{-inb}-e^{-ina})$ .

This is my approach:
$\sum _{n\neq 0}\left|e^{inx}{\frac {1}{2\pi in}}(e^{-inb}-e^{-ina})\right|$
$=\sum _{n\neq 0}{\frac {1}{2\pi n}}\left|(e^{-inb}-e^{-ina})\right|$
This is almost the harmonic series. The desired result follows if you can show that $\left|(e^{-inb}-e^{-ina})\right|>c$ for some $c>0$ for every $n$ in an arithmetic progression, for example. Equivalently,
$\left|(e^{-inb}-e^{-ina})\right|>c$
$\iff 1\left|(e^{-inb}-e^{-ina})\right|>c$
$\iff |e^{ina}|\left|(e^{-inb}-e^{-ina})\right|>c$
$\iff \left|(e^{in(a-b)}-1)\right|>c$
might be easier to show.
$\iff \left|(\cos(n(a-b))+i\sin(n(a-b))-1)\right|>c$
Then it suffices that $\sin(n(a-b))>d$ for all $0<d<1$ for all $n$ in an arithmetic progression which is why I asked a similar question a while ago. That way, the number inside the absolute value signs will always have a nonzero imaginary part and therefore will have nonzero absolute value. It's not actually enough that $\sin(n(a-b))>d$ for infinitely many $n$ as I later realized, for instance the result doesn't (at least not obviously) follow if its only true for the squares.
Anyway, do you have suggestions for proving this result? — Preceding unsigned comment added by Widener (talk • contribs) 20:07, 19 September 2012 (UTC)[reply]

exp(-i n b) - exp(- i n a) =

exp[-i n (a+b)/2] {exp(i n (a-b)/2) - exp(-i n (a-b)/2)}

So, the summation of the absolute value of the terms is:

{\frac {2}{\pi }}\sum _{n=1}^{\infty }{\frac {\left|\sin \left[{\frac {(a-b)n}{2}}\right]\right|}{n}}

If you replace infinity by N in the upper limit, you can say that the value is larger than what you get if you replace the absolute value of sin by sin^2. Then you can write sin^2 in terms of cos of the double argument, this yields one divergent summation (in the limit of N to infinity) and a convergent summation. Count Iblis (talk) 20:55, 19 September 2012 (UTC)[reply]

Hey, that works! Thanks! Widener (talk) 21:16, 19 September 2012 (UTC)[reply]

The maze on today's main page.

As I posted earlier on Wikipedia:Main Page/Errors, the caption for this was a little confusing. Can anyone here figure out what it was trying to say: "In these mazes, it will typically be relatively easy to find the origin point, since most paths lead to or from there, but it is hard to find the way out"? As far as I can tell, since everywhere is connected to everywhere else, and there is neither a marked 'start' or 'end' once the maze is constructed, the statement makes no sense.

Incidentally, I'm fairly sure that there is a name for this particular type of maze - it has no 'loops' so can be navigated in its entirety simply by 'following a wall' - does anyone know the name? AndyTheGrump (talk) 21:08, 19 September 2012 (UTC)[reply]

It sounds like the "all roads lead to Rome" concept. In the case of the Roman Empire, all roads would actually also lead to any connected city. What they are really saying is that it's the central hub. Think of it like an airline's hub city, too. Yes, you can get to any city they service eventually, but the direct routes all lead to or from the hub city. StuRat (talk) 00:34, 20 September 2012 (UTC)[reply]

The phrase "simply connected" comes to mind, and it would technically be correct for a certain view of this (no loops) maze, although I'm not sure if anyone in the maze community uses precisely that term. (Although simple graph would be more relevant from a graph theory perspective.) I would also imagine (though don't have any proof/calculations to show) that the depth-first generation method would produce a maze with a low average degree of branching, which should aid in maze traversal, though I don't know if it would differentially affect and given position. Although it might be that since branching preferentially happens at the tips of the growing maze, you might expect a gradient of degree, with points near the origin having lower average degree than those further from it, leading to search being simpler in the region of the starting point. -- 205.175.124.30 (talk) 02:11, 20 September 2012 (UTC)[reply]

It appears that "perfect maze" is the most common name for loop-free mazes. -- BenRG (talk) 06:10, 20 September 2012 (UTC)[reply]

Why you force SSL in the HTTP link...? http://www.google.com/search?q=perfect+maze
See also directly pages Maze and Maze solving algorithm, they both say the same: Mazes containing no loops are known as "standard", or "perfect" mazes, and are equivalent to a tree in graph theory. --CiaPan (talk) 05:23, 21 September 2012 (UTC)[reply]

Why you force unsecure plaintext in the HTTP link? Nil Einne (talk) 13:32, 25 September 2012 (UTC)[reply]

Beacuse it doesn't need any 'security' and there's no reason to hide it. There's nothing secret in using Google search machine or in the query itself or in its results. Additionally pasting the plain URL makes it easier for readers to see where they are directed. It is also short enough, so it does not disturb in reading surrounding text. --CiaPan (talk) 15:25, 25 September 2012 (UTC)[reply]