Cantor's intersection theorem

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

Theorem 2: If ${\displaystyle X}$ is a compact space and ${\displaystyle \{C_{n}\}}$ is an infinite sequence of non-empty, closed sets such that ${\displaystyle C_{1}\supseteq \cdots C_{k}\supseteq C_{k+1}\cdots ,\,}$, then ${\displaystyle \bigcap _{n=1}^{\infty }C_{n}\neq \varnothing }$.

Notice the differences and the similarities between the two theorem. In Theorem 2, the ${\displaystyle C_{n}}$ are only required to be closed since given a compact space ${\displaystyle X}$ and ${\displaystyle Y\subset X}$ a closed subset, then ${\displaystyle Y}$ 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 can only have the Cantor Intersection property (i.e. the theorem above holds) if it is complete (for justification see below). An example of an application of this theorem is the existence of limit points for self-similar contracting fractals^[2].

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. ${\displaystyle \mathbb {Q} }$ with the usual metric and the sequence of sets, ${\displaystyle C_{n}=[{\sqrt {2}},{\sqrt {2}}+1/n]}$. If the sets are not closed, then one can construct sequences of nested sets which have empty intersection, i.e. ${\displaystyle \mathbb {R} }$ with the collection, ${\displaystyle C_{n}=(0-{\frac {1}{n}},0+{\frac {1}{n}})}$ or the collection ${\displaystyle C_{n}=[n,\infty )}$. This last case also demonstrates what can happen if the diameters are not tending to 0. However, another example would be ${\displaystyle C_{n}=[0-{\frac {1}{n}},1+{\frac {1}{n}}]}$ and the infinite intersection will yield more than a single point. Violation of the remaining hypotheses are clear and left to the reader as an exercise (I have always wanted to write that).

Proof

Theorem 1: Suppose ${\displaystyle (X,d)}$  is a non-trivial, complete metric space and ${\displaystyle \{C_{n}\}}$  is an infinite family of non-empty closed sets in ${\displaystyle X}$  such that ${\displaystyle C_{n}\supset C_{n+1},\forall n}$  and ${\displaystyle \lim _{n\to \infty }diam(C_{n})\rightarrow 0}$ . Naturally we would like to use the completeness so we will construct a Cauchy sequence. Since each of the ${\displaystyle C_{n}}$  is closed, there exists a ${\displaystyle y_{n}}$  in the interior (i.e. positive distance to anything outside ${\displaystyle C_{n}}$ ) of ${\displaystyle C_{n}}$ . These ${\displaystyle y_{n}}$  form a sequence. Since ${\displaystyle \lim _{n\to \infty }diam(C_{n})\rightarrow 0}$ , then given any positive real value, ${\displaystyle \epsilon >0}$ , there exists a large ${\displaystyle N}$  such that whenever ${\displaystyle n\geq N}$ , ${\displaystyle diam(C_{n})<\epsilon }$ . Since, ${\displaystyle C_{n}\supset C_{n+1},\forall n}$ , then given any ${\displaystyle n,m\geq N}$ , ${\displaystyle y_{n},y_{m}\in C_{n}}$  and therefore, ${\displaystyle d(y_{n},y_{m})<\epsilon }$ . Thus, the ${\displaystyle y_{n}}$  form a Cauchy sequence. By the completeness of ${\displaystyle X}$  there is a point ${\displaystyle x\in X}$  such that ${\displaystyle y_{n}\to x}$ . By the closure of each ${\displaystyle C_{n}}$  and since ${\displaystyle x}$  is in ${\displaystyle C_{n}}$  for all ${\displaystyle n\geq N}$ , ${\displaystyle x\in \bigcap _{n=1}^{\infty }C_{n}}$ . To see that ${\displaystyle x}$  is alone in ${\displaystyle \bigcap _{n=1}^{\infty }C_{n}}$  assume otherwise. Take ${\displaystyle x'\in \bigcap _{n=1}^{\infty }C_{n}}$  and then consider the distance between ${\displaystyle x}$  and ${\displaystyle x'}$  this is some value greater than 0 and implies that the ${\displaystyle \lim _{n\to \infty }diam(C_{n})\rightarrow d(x,x')>0}$ . Contradiction! Thus, ${\displaystyle x}$  is very, very lonely in his small spartan little dorm room that is the infinite intersection of a sequence of closed set in a metric space that is complete, but how would he ever know.

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

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

References

1. "Real Analysis," H.L. Royden, P.M. Fitzpatrick, 4th edition, 2010, page 195
2. Ergodic Theory and Symbolic Dynamics in Hyperbolic Spaces, T. Bedford, M. Keane and C. Series eds., Oxford Univ. Press 1991, page 225

• Weisstein, Eric W. "Cantor's Intersection Theorem". MathWorld.
• Jonathan Lewin. An interactive introduction to mathematical analysis. Cambridge University Press. ISBN 0-521-01718-1. Section 7.8.