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== 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.
'''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 compliment 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> since I am throwing out less and less stuff each time. 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>.
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