Conjectural variation - Revision history

66.74.176.59: sp

2014-11-18T08:41:29Z

← Older revision		Revision as of 10:41, 18 November 2014
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	~~{{Underlinked\|date=September 2013}}~~		Hi, everybody! My name is Robyn. <br>It is a little about myself: I live in Netherlands, my city of Burgh-Haamstede. <br>It's called often Eastern or cultural capital of ZE. I've married 2 years ago.<br>I have 2 children - a son (Kelsey) and the daughter (Kala). We all like Association football.<br><br>My web blog :: [http://jatekok.jatekok.name/profile/lathorp Choosing the right ride for you mountain bike sizing.]

	~~In [[real analysis]]~~, ~~a branch of mathematics, '''Cantor's intersection theorem''', named after [[Georg Cantor]], gives conditions under which an infinite intersection of nested, non-empty, sets~~ is ~~non-empty~~.

	~~'''Theorem 1''': If~~ <~~math>(X, d)</math~~> is a ~~non-trivial~~, ~~[[complete metric space]] and <math>\{C_n\}</math> is an infinite sequence~~ of ~~non~~-~~empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\rightarrow\infty} diam(C_n)=\sup\{d(x,y): x,y\in C_n\}\rightarrow 0</math>~~. ~~Then, there exists an~~ <~~math~~>~~x\in X</math> such that <math>\bigcap_{n=1}^\infty C_n = \{x\} </math>.<ref>"Real Analysis," H.L. Royden, P.M. Fitzpatrick, 4th edition, 2010, page 195</ref>~~

	'~~''Theorem 2''': If <math>X</math> is a [[compact (mathematics) \| compact]] space and <math>\{C_n\}</math> is an infinite sequence~~ of ~~non-empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math>, then <math>\bigcap_{n=1}^\infty C_n\neq\varnothing</math>~~.

	~~Notice the differences and the similarities between the two theorem. In Theorem~~ 2, the <math>C_n</math> are only assumed to be closed (and not compact, which is stronger) since given a compact space <math>X</math> and <math>Y\subset X</math> a closed subset, <math>Y</math> is necessarily compact. Also, in Theorem 1 the intersection is exactly 1 point, while in Theorem 2 it could contain many more points. Interestingly, a metric space having the Cantor Intersection property (i.e. the theorem above holds) is necessarily complete (for justification see below). An example of an application of this theorem is the existence of limit points for self-similar contracting fractals.<~~ref~~>~~Ergodic Theory and Symbolic Dynamics in Hyperbolic Spaces, T. Bedford, M. Keane and C. Series eds., Oxford Univ. Press 1991, page 225</ref>~~

	Notice that each of the hypotheses above is essential. If the metric space were not complete, then one could construct a nested sequence of non-empty, compact sets converging to a "hole" in the space, i.e. <math>\mathbb{Q}</math> with the usual metric and the sequence of sets, <math>C_n = [\sqrt{2}, \sqrt{2}+1/n]</math>. If the sets are not closed, then one can construct sequences of nested sets which have empty intersection, i.e. <math>\mathbb{R}</math> with the collection, <math>C_n = (0,\frac{1}{n}) </math>. The collections <math>C_n = [n, \infty)</math> and <math>C_n = [-~~\frac{1}{n}, 1+\frac{1}{n}]</math> illustrate what may happen when the diameters do not tend to zero: the intersection may be empty, as in the first, or may contain more than~~ a ~~single point, as in the second.~~

	~~== Proof ==~~

	~~'''Theorem 1''':~~
	~~Suppose <math>~~(~~X,d~~)~~</math> is a non-trivial, complete metric space~~ and <math>\{C_n\}</math> is an infinite family of non-empty closed sets in <math>X</math> such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\to\infty} diam(C_n)\rightarrow 0</math>. Naturally we would like to use the ~~completeness so we will construct a Cauchy sequence. Since each of the <math>C_n</math> is closed, there exists a <math>y_n</math> in the interior~~ (~~i.e. positive distance to anything outside <math>C_n</math>~~) ~~of <math>C_n</math>~~. These <math>y_n</math> form a sequence. Since <math>\lim_{n\to\infty} diam(C_n)\rightarrow 0</math>, then given any positive real value, <math>\epsilon>0</math>, there exists a large <math>N</math> such that whenever <math>n\geq N</math>, <math>diam(C_n)<\epsilon</math>. Since, <math>C_n\supset C_{n+1},\forall n</math>, then given any <math>n,m\geq N</math>, <math>y_n,y_m \in C_n</math> and therefore, <math>d(y_n,y_m)<\epsilon</math>. Thus, the <math>y_n</math> form a Cauchy sequence. By the completeness of <math>X</math> there is a point <math>x\in X</math> such that <math>y_n\to x</math>. By the closure of each <math>C_n</math> and since <math>x</math> is in <math>C_n</math> for all ~~<math>n\geq N</math>, <math>x\in\bigcap_{n=1}^\infty C_n</math>~~. ~~To see that <math>x~~<~~/math~~> ~~is alone in~~ <~~math~~>\bigcap_{n=1}^\infty C_n</math> assume otherwise. Take <math>x'\in\bigcap_{n=1}^\infty C_n</math> and then consider the distance between <math>x</math> and <math>x'</math> this is some value greater than 0 and implies that the <math>\lim_{n\to\infty} diam(C_n)\rightarrow d(x,x')>0</math>. Contradiction! Thus the claim follows.

	~~'''Theorem 2'''~~:
	Suppose <math>X</math> is a compact topological space and <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_1 \supseteq \cdots C_k \supseteq C_{k+1} \cdots, \, </math>. Assume, by contradiction, that <math>\bigcap_{n=1}^\infty C_n=\varnothing</math>. Then we will build an open cover of <math>X</math> by considering the complement of <math>C_n</math> in <math>X</math>, i.e. <math>U_n=X\setminus C_n,\forall n</math>. Each <math>U_n</math> is open since the <math>C_n</math> are closed. Notice that <math>\bigcup_{n=1}^\infty U_n = \bigcup_{n=1}^\infty (X\setminus C_n) = X\setminus\bigcap_{n=1}^\infty C_n</math>, but we assumed that <math>\bigcap_{n=1}^\infty C_n=\varnothing</math> so that means <math>\bigcup_{n=1}^\infty U_n = X</math>. So, there are infinite many <math>U_n</math> covering our compact <math>X</math>. That means there exists a large <math>N</math> such that <math> X\subset\bigcup_{n=1}^N U_n</math>. Notice, however, that <math>C_n\supset C_{n+1},\forall n</math> implies that <math>X\setminus C_n=U_n\subset U_{n+1}=X\setminus C_{n+1},\forall n</math>. The only way for the nested and increasing <math>U_n</math> to cover <math>X</math> is if there is some index, call it <math>k</math>, such that <math>X=U_k</math>. This implies though that <math>C_k=X\setminus U_k=\varnothing</math>. This is a contradiction since we assumed that the <math>C_n</math> were non-empty. Hence, <math>\bigcap_{n=1}^\infty C_n\neq\varnothing</math>.

	~~Notice that in regards to the proof of Theorem 2, we don't need Hausdorffness. At no point in time do we appeal to the nature of points in the space. It is simply a statement about empty or not.~~

	Consider now a metric space <math>(X,d)</math> (not necessarily complete) in which <math>\bigcap_{n=1}^\infty C_n = x </math> whenever <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\to\infty} diam(C_n)=sup\{d(x,y): ~~x,y\in X\}\rightarrow 0</math>. Now, let <math>\{x_k\}</math> be a Cauchy sequence in <math> X</math> and take <math>C_n=\overline{\{x_k~~:k\geq n\}}</math>. The bar over the set means that we are taking the closure of the set under it. This guarantees that we are working with closed sets and since they contain the elements of our Cauchy sequence, we know them to be non-empty. In addition, <math>C_n\supset C_{n+1}</math> and since <math>\forall\epsilon>0,\exists N</math> such that when <math>n,m\geq N,d(x_n,x_m)<\epsilon</math>, (note this hold for all indices larger than our large <math>N</~~math>) then <math>diam(C_N)<\epsilon<~~/~~math>~~. ~~Hence, <math>\{C_n\}</math> satisfies the conditions above and there exists an <math>x\in X</math> such that <math>\bigcap_{n=1}^\infty C_n = x </math>~~. ~~So, <math>x<~~/~~math> is in the closure of all of the <math>C_n</math> and any open ball around <math>x<~~/~~math> has non-empty intersection with~~ the ~~<math>C_n</math>~~. Now we will build a sub-sequence of the <math>\{x_n\}</math>, call it <math>\{x_{n_k}\}</math>, where <math>d(x,x_{n_k})<\frac{1}{k}</math>. This implies that <math>\{x_{n_k}\}\to x</math> and since <math>\{x_k\}</math> was Cauchy then it too must converge to <math>x</math>. Since <math>\{x_k\}</math> was an arbitrary Cauchy sequence, <math>X</math> is complete.

	~~== References ==~~
	~~{{Reflist}}~~
	* {{MathWorld \| urlname=CantorsIntersectionTheorem \| title=Cantor's Intersection Theorem}}
	* Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.

	~~[[Category:Articles containing proofs]]~~
	~~[[Category:Real analysis]]~~
	~~[[Category:Compactness theorems]]~~
	~~[[Category:Theorems in calculus]~~]

131.251.133.28: /* Consistent conjectures */ spelling

2013-07-31T15:41:28Z

Consistent conjectures: spelling

← Older revision		Revision as of 17:41, 31 July 2013
Line 1:		Line 1:
	~~Make sure you can hear your breathe~~ (~~except~~ in ~~public places where you may want~~ to be ~~more discreet~~) and ~~you can see~~ the ~~stomach moving as this will help you to build up a rhythm~~. ~~If you have toe clips on your pedals~~, ~~you'll need to spend~~ a ~~little extra time with~~ this ~~exercise~~. ~~The downhill bikes have both front~~ and ~~rear suspension~~, ~~disc breaks~~ and ~~are very strong~~. If the ~~skin diameter is smaller than a quarter do~~ not ~~be surprised if you must change them~~ a ~~couple~~ of ~~instances per season~~ to ~~keep~~ the ~~bike working reliably~~. ~~There~~ are ~~many times you will need to grip~~ the ~~handle of your bike tightly when you negotiate difficult trails~~. <br><br>~~You can choose from many different suspensions on your bike~~, ~~make sure~~ the ~~suspension you choose is going~~ to ~~fit~~ the ~~type of cycling you intend for it. In fact~~, ~~Downhill Mountain biking is~~ the ~~most popular form of competition biking. You can spend under $100 for a bargain bike at a department store~~, or ~~lay down thousands for~~ a ~~professional model~~. ~~If you liked this article~~ and ~~you also~~ would like to ~~acquire more info pertaining~~ to ~~[http:~~//~~www~~.~~isaacwkim~~.~~com~~/xe/~~?document_srl=944635 Popular mountain bike sizing~~.~~] please visit~~ the ~~web site~~. ~~Ten miles back to~~ the ~~truck~~ is a ~~long walk when pushing 200 pounds~~ of ~~meat on a bike~~. ~~Titanium~~ is ~~very light~~ and ~~gives you an advantage~~ and is ~~usually used in racing bicycles~~. <br><br>~~It seems there~~ is ~~still ongoing trail expansion~~, ~~so I predict a great future for the Gold Canyon trails~~. ~~People now prefer to have~~ the ~~simplest design~~ in ~~their bikes as long as it can get~~ the ~~job well done~~. ~~Buy~~ a ~~bike~~ that ~~has a top quality body made outside of steel~~, ~~aluminum~~, ~~carbon fiber or titanium~~. ~~Hence~~, ~~having the bicycle~~ that ~~suits you perfectly~~ is ~~much easier now. You might want to pitch~~ a ~~tent or you may find~~ that ~~there are many other options for lodging as you move along~~ the ~~trails on your mountain biking trips~~. <br><br>~~There are still other types of bikes~~ that ~~require different types~~ of ~~tires~~. ~~This article discusses how~~ to go about ~~buying~~ a ~~mountain bike~~. ~~Serious head injuries can~~ be ~~prevented with~~ a ~~helmet~~ and ~~today’s styles and designs make them more comfortable to wear~~. ~~I should have had~~ that ~~Bicycle repair tool~~ with ~~me, becuase when I got home~~ and ~~lowered~~ the ~~seat~~. ~~I've never been into cycling~~ and ~~wanted to give it a try~~ for ~~exercise~~. <br><br>In the ~~recent article section~~, ~~you can find~~ the ~~list~~ of ~~biking related articles~~ with a ~~detailed description present on~~ the ~~official website. There is anything from shocks to gears~~, ~~special wheels~~, ~~exclusive take care of bars~~ that ~~can be immediately switched~~ to ~~match the terrain you're riding on~~ and ~~so substantially much more~~. ~~Trying out various sizes~~ is ~~the first step in choosing the correct folding bike~~. ~~Sign up~~ to ~~get a mountain bike expedition and relive people childhood biking wonder and thrills with a whole new tier~~. ~~Part of Park City's distinction as the International Mountain Bike Association's first and only gold~~-~~level Ride Center is its abundance of beginner trails~~.		{{Underlinked\|date=September 2013}}

			In [[real analysis]], a branch of mathematics, '''Cantor's intersection theorem''', named after [[Georg Cantor]], gives conditions under which an infinite intersection of nested, non-empty, sets is non-empty.

			'''Theorem 1''': If <math>(X, d)</math> is a non-trivial, [[complete metric space]] and <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\rightarrow\infty} diam(C_n)=\sup\{d(x,y): x,y\in C_n\}\rightarrow 0</math>. Then, there exists an <math>x\in X</math> such that <math>\bigcap_{n=1}^\infty C_n = \{x\} </math>.<ref>"Real Analysis," H.L. Royden, P.M. Fitzpatrick, 4th edition, 2010, page 195</ref>

			'''Theorem 2''': If <math>X</math> is a [[compact (mathematics) \| compact]] space and <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math>, then <math>\bigcap_{n=1}^\infty C_n\neq\varnothing</math>.

			Notice the differences and the similarities between the two theorem. In Theorem 2, the <math>C_n</math> are only assumed to be closed (and not compact, which is stronger) since given a compact space <math>X</math> and <math>Y\subset X</math> a closed subset, <math>Y</math> is necessarily compact. Also, in Theorem 1 the intersection is exactly 1 point, while in Theorem 2 it could contain many more points. Interestingly, a metric space having the Cantor Intersection property (i.e. the theorem above holds) is necessarily complete (for justification see below). An example of an application of this theorem is the existence of limit points for self-similar contracting fractals.<ref>Ergodic Theory and Symbolic Dynamics in Hyperbolic Spaces, T. Bedford, M. Keane and C. Series eds., Oxford Univ. Press 1991, page 225</ref>

			Notice that each of the hypotheses above is essential. If the metric space were not complete, then one could construct a nested sequence of non-empty, compact sets converging to a "hole" in the space, i.e. <math>\mathbb{Q}</math> with the usual metric and the sequence of sets, <math>C_n = [\sqrt{2}, \sqrt{2}+1/n]</math>. If the sets are not closed, then one can construct sequences of nested sets which have empty intersection, i.e. <math>\mathbb{R}</math> with the collection, <math>C_n = (0,\frac{1}{n}) </math>. The collections <math>C_n = [n, \infty)</math> and <math>C_n = [-\frac{1}{n}, 1+\frac{1}{n}]</math> illustrate what may happen when the diameters do not tend to zero: the intersection may be empty, as in the first, or may contain more than a single point, as in the second.

			== Proof ==

			'''Theorem 1''':
			Suppose <math>(X,d)</math> is a non-trivial, complete metric space and <math>\{C_n\}</math> is an infinite family of non-empty closed sets in <math>X</math> such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\to\infty} diam(C_n)\rightarrow 0</math>. Naturally we would like to use the completeness so we will construct a Cauchy sequence. Since each of the <math>C_n</math> is closed, there exists a <math>y_n</math> in the interior (i.e. positive distance to anything outside <math>C_n</math>) of <math>C_n</math>. These <math>y_n</math> form a sequence. Since <math>\lim_{n\to\infty} diam(C_n)\rightarrow 0</math>, then given any positive real value, <math>\epsilon>0</math>, there exists a large <math>N</math> such that whenever <math>n\geq N</math>, <math>diam(C_n)<\epsilon</math>. Since, <math>C_n\supset C_{n+1},\forall n</math>, then given any <math>n,m\geq N</math>, <math>y_n,y_m \in C_n</math> and therefore, <math>d(y_n,y_m)<\epsilon</math>. Thus, the <math>y_n</math> form a Cauchy sequence. By the completeness of <math>X</math> there is a point <math>x\in X</math> such that <math>y_n\to x</math>. By the closure of each <math>C_n</math> and since <math>x</math> is in <math>C_n</math> for all <math>n\geq N</math>, <math>x\in\bigcap_{n=1}^\infty C_n</math>. To see that <math>x</math> is alone in <math>\bigcap_{n=1}^\infty C_n</math> assume otherwise. Take <math>x'\in\bigcap_{n=1}^\infty C_n</math> and then consider the distance between <math>x</math> and <math>x'</math> this is some value greater than 0 and implies that the <math>\lim_{n\to\infty} diam(C_n)\rightarrow d(x,x')>0</math>. Contradiction! Thus the claim follows.

			'''Theorem 2''':
			Suppose <math>X</math> is a compact topological space and <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_1 \supseteq \cdots C_k \supseteq C_{k+1} \cdots, \, </math>. Assume, by contradiction, that <math>\bigcap_{n=1}^\infty C_n=\varnothing</math>. Then we will build an open cover of <math>X</math> by considering the complement of <math>C_n</math> in <math>X</math>, i.e. <math>U_n=X\setminus C_n,\forall n</math>. Each <math>U_n</math> is open since the <math>C_n</math> are closed. Notice that <math>\bigcup_{n=1}^\infty U_n = \bigcup_{n=1}^\infty (X\setminus C_n) = X\setminus\bigcap_{n=1}^\infty C_n</math>, but we assumed that <math>\bigcap_{n=1}^\infty C_n=\varnothing</math> so that means <math>\bigcup_{n=1}^\infty U_n = X</math>. So, there are infinite many <math>U_n</math> covering our compact <math>X</math>. That means there exists a large <math>N</math> such that <math> X\subset\bigcup_{n=1}^N U_n</math>. Notice, however, that <math>C_n\supset C_{n+1},\forall n</math> implies that <math>X\setminus C_n=U_n\subset U_{n+1}=X\setminus C_{n+1},\forall n</math>. The only way for the nested and increasing <math>U_n</math> to cover <math>X</math> is if there is some index, call it <math>k</math>, such that <math>X=U_k</math>. This implies though that <math>C_k=X\setminus U_k=\varnothing</math>. This is a contradiction since we assumed that the <math>C_n</math> were non-empty. Hence, <math>\bigcap_{n=1}^\infty C_n\neq\varnothing</math>.

			Notice that in regards to the proof of Theorem 2, we don't need Hausdorffness. At no point in time do we appeal to the nature of points in the space. It is simply a statement about empty or not.

			Consider now a metric space <math>(X,d)</math> (not necessarily complete) in which <math>\bigcap_{n=1}^\infty C_n = x </math> whenever <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_n\supset C_{n+1},\forall n</math> and <math>\lim_{n\to\infty} diam(C_n)=sup\{d(x,y): x,y\in X\}\rightarrow 0</math>. Now, let <math>\{x_k\}</math> be a Cauchy sequence in <math> X</math> and take <math>C_n=\overline{\{x_k:k\geq n\}}</math>. The bar over the set means that we are taking the closure of the set under it. This guarantees that we are working with closed sets and since they contain the elements of our Cauchy sequence, we know them to be non-empty. In addition, <math>C_n\supset C_{n+1}</math> and since <math>\forall\epsilon>0,\exists N</math> such that when <math>n,m\geq N,d(x_n,x_m)<\epsilon</math>, (note this hold for all indices larger than our large <math>N</math>) then <math>diam(C_N)<\epsilon</math>. Hence, <math>\{C_n\}</math> satisfies the conditions above and there exists an <math>x\in X</math> such that <math>\bigcap_{n=1}^\infty C_n = x </math>. So, <math>x</math> is in the closure of all of the <math>C_n</math> and any open ball around <math>x</math> has non-empty intersection with the <math>C_n</math>. Now we will build a sub-sequence of the <math>\{x_n\}</math>, call it <math>\{x_{n_k}\}</math>, where <math>d(x,x_{n_k})<\frac{1}{k}</math>. This implies that <math>\{x_{n_k}\}\to x</math> and since <math>\{x_k\}</math> was Cauchy then it too must converge to <math>x</math>. Since <math>\{x_k\}</math> was an arbitrary Cauchy sequence, <math>X</math> is complete.

			== References ==
			{{Reflist}}
			* {{MathWorld \| urlname=CantorsIntersectionTheorem \| title=Cantor's Intersection Theorem}}
			* Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.

			[[Category:Articles containing proofs]]
			[[Category:Real analysis]]
			[[Category:Compactness theorems]]
			[[Category:Theorems in calculus]]

en>Magioladitis: Fix category spacing + general fixes using AWB (8232)

2012-08-07T08:25:17Z

Fix category spacing + general fixes using AWB (8232)

New page

Make sure you can hear your breathe (except in public places where you may want to be more discreet) and you can see the stomach moving as this will help you to build up a rhythm. If you have toe clips on your pedals, you'll need to spend a little extra time with this exercise. The downhill bikes have both front and rear suspension, disc breaks and are very strong. If the skin diameter is smaller than a quarter do not be surprised if you must change them a couple of instances per season to keep the bike working reliably. There are many times you will need to grip the handle of your bike tightly when you negotiate difficult trails. You can choose from many different suspensions on your bike, make sure the suspension you choose is going to fit the type of cycling you intend for it. In fact, Downhill Mountain biking is the most popular form of competition biking. You can spend under $100 for a bargain bike at a department store, or lay down thousands for a professional model. If you liked this article and you also would like to acquire more info pertaining to [http://www.isaacwkim.com/xe/?document_srl=944635 Popular mountain bike sizing.] please visit the web site. Ten miles back to the truck is a long walk when pushing 200 pounds of meat on a bike. Titanium is very light and gives you an advantage and is usually used in racing bicycles. It seems there is still ongoing trail expansion, so I predict a great future for the Gold Canyon trails. People now prefer to have the simplest design in their bikes as long as it can get the job well done. Buy a bike that has a top quality body made outside of steel, aluminum, carbon fiber or titanium. Hence, having the bicycle that suits you perfectly is much easier now. You might want to pitch a tent or you may find that there are many other options for lodging as you move along the trails on your mountain biking trips. There are still other types of bikes that require different types of tires. This article discusses how to go about buying a mountain bike. Serious head injuries can be prevented with a helmet and today’s styles and designs make them more comfortable to wear. I should have had that Bicycle repair tool with me, becuase when I got home and lowered the seat. I've never been into cycling and wanted to give it a try for exercise. In the recent article section, you can find the list of biking related articles with a detailed description present on the official website. There is anything from shocks to gears, special wheels, exclusive take care of bars that can be immediately switched to match the terrain you're riding on and so substantially much more. Trying out various sizes is the first step in choosing the correct folding bike. Sign up to get a mountain bike expedition and relive people childhood biking wonder and thrills with a whole new tier. Part of Park City's distinction as the International Mountain Bike Association's first and only gold-level Ride Center is its abundance of beginner trails.