Content deleted Content added
No edit summary |
|||
Line 14:
'''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
'''Theorem 2''':
| |||