Cantor's intersection theorem: Difference between revisions

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'''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\to\infty} diam(C_n)=sup\{d(x,y): x,y\in X\}\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 space and <math>\{C_n\}</math> is an infinite sequence of non-empty, closed sets such that <math>C_1 >C_n\supseteq \cdots C_k \supseteqsupset C_{kn+1} \cdots, \,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 requiredassumed 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, then <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 can only havehaving the Cantor Intersection property (i.e. the theorem above holds) ifis it isnecessarily 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},0+\frac{1}{n}) </math> or the collection <math>C_n = [n, \infty)</math>. This last case also demonstrates what can happen if the diameters are not tending to 0. However, another example would be <math>C_n = [0-\frac{1}{n}, 1+\frac{1}{n}]</math> 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 <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, <math>x</math> 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.