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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>
Notice the differences and the similarities between the two theorem. In Theorem 2, the <math>C_n</math> are only
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
== 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.
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