Conjectural variation

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 $(X,d)$ is a non-trivial, complete metric space and $\{C_n\}$ is an infinite sequence of non-empty, closed sets such that $C_n\supset C_{n+1},\mathrm{\forall }n$ and $\underset{n\to \mathrm{\infty }}{\lim }diam(C_n)=\sup \{d(x,y):x,y\in C_n\}\to 0$. Then, there exists an $x\in X$ such that $\bigcap _{n=1}^{\mathrm{\infty }}{C}_n=\{x\}$.

Theorem 2: If $X$ is a compact space and $\{C_n\}$ is an infinite sequence of non-empty, closed sets such that $C_n\supset C_{n+1},\mathrm{\forall }n$, then $\bigcap _{n=1}^{\mathrm{\infty }}{C}_n\ne \varnothing$.

Notice the differences and the similarities between the two theorem. In Theorem 2, the $C_n$ are only assumed to be closed (and not compact, which is stronger) since given a compact space $X$ and $Y\subset X$ a closed subset, $Y$ 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.

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. $\mathbb{\mathbb{Q}}$ with the usual metric and the sequence of sets, $C_n=[\sqrt{2},\sqrt{2}+1/n]$. If the sets are not closed, then one can construct sequences of nested sets which have empty intersection, i.e. $\mathbb{ℝ}$ with the collection, $C_n=(0,\frac{1}{n})$. The collections $C_n=[n,\mathrm{\infty })$ and $C_n=[-\frac{1}{n},1+\frac{1}{n}]$ 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 $(X,d)$ is a non-trivial, complete metric space and $\{C_n\}$ is an infinite family of non-empty closed sets in $X$ such that $C_n\supset C_{n+1},\mathrm{\forall }n$ and $\underset{n\to \mathrm{\infty }}{\lim }diam(C_n)\to 0$. Naturally we would like to use the completeness so we will construct a Cauchy sequence. Since each of the $C_n$ is closed, there exists a $y_n$ in the interior (i.e. positive distance to anything outside $C_n$) of $C_n$. These $y_n$ form a sequence. Since $\underset{n\to \mathrm{\infty }}{\lim }diam(C_n)\to 0$, then given any positive real value, $ϵ>0$, there exists a large $N$ such that whenever $n\ge N$, $diam(C_n)<ϵ$. Since, $C_n\supset C_{n+1},\mathrm{\forall }n$, then given any $n,m\ge N$, $y_n,y_m\in C_n$ and therefore, $d(y_n,y_m)<ϵ$. Thus, the $y_n$ form a Cauchy sequence. By the completeness of $X$ there is a point $x\in X$ such that $y_n\to x$. By the closure of each $C_n$ and since $x$ is in $C_n$ for all $n\ge N$, $x\in \bigcap _{n=1}^{\mathrm{\infty }}{C}_n$. To see that $x$ is alone in $\bigcap _{n=1}^{\mathrm{\infty }}{C}_n$ assume otherwise. Take $x^{\prime }\in \bigcap _{n=1}^{\mathrm{\infty }}{C}_n$ and then consider the distance between $x$ and $x^{\prime }$ this is some value greater than 0 and implies that the $\underset{n\to \mathrm{\infty }}{\lim }diam(C_n)\to d(x,x^{\prime })>0$. Contradiction! Thus the claim follows.

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

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 $(X,d)$ (not necessarily complete) in which $\bigcap _{n=1}^{\mathrm{\infty }}{C}_n=x$ whenever $\{C_n\}$ is an infinite sequence of non-empty, closed sets such that $C_n\supset C_{n+1},\mathrm{\forall }n$ and $\underset{n\to \mathrm{\infty }}{\lim }diam(C_n)=sup\{d(x,y):x,y\in X\}\to 0$. Now, let $\{x_k\}$ be a Cauchy sequence in $X$ and take $C_n=\overline{\{x_k:k\ge n\}}$. 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, $C_n\supset C_{n+1}$ and since $\mathrm{\forall }ϵ>0,\mathrm{\exists }N$ such that when $n,m\ge N,d(x_n,x_m)<ϵ$, (note this hold for all indices larger than our large $N$) then $diam(C_N)<ϵ$. Hence, $\{C_n\}$ satisfies the conditions above and there exists an $x\in X$ such that $\bigcap _{n=1}^{\mathrm{\infty }}{C}_n=x$. So, $x$ is in the closure of all of the $C_n$ and any open ball around $x$ has non-empty intersection with the $C_n$. Now we will build a sub-sequence of the $\{x_n\}$, call it $\{x_{n_k}\}$, where $d(x,x_{n_k})<\frac{1}{k}$. This implies that $\{x_{n_k}\}\to x$ and since $\{x_k\}$ was Cauchy then it too must converge to $x$. Since $\{x_k\}$ was an arbitrary Cauchy sequence, $X$ is complete.

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

• 
  
  I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
  
  
  
  Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.
• Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.