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''for some <math>x \in X</math>.''
''Proof (sketch).'' Since the diameters tend to zero, the diameter of the intersection of the <math>C_k</math> is zero, so it is either empty or consists of a single point. So it is sufficient to show that it is not empty. Pick an element <math>x_k\in C_k</math> for each <math>k</math>. Since the diameter of <math>C_k</math> tends to zero and the <math>C_k</math> are nested, the <math>x_k</math> form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point <math>x</math>. Since each <math>C_k</math> is closed, and <math>x</math> is a limit of a sequence in <math>C_k</math>, <math>x</math> must lie in <math>C_k</math>. This is true for every <math>k</math>, and therefore the intersection of the <math>C_k</math> must contain <
A converse to this theorem is also true: if <math>X</math> is a metric space with the property that the intersection of any nested family of non-empty closed subsets whose diameters tend to zero is non-empty, then <math>X</math> is a complete metric space. (To prove this, let <math>(x_k)_{k \geq 1}</math> be a Cauchy sequence in <math>X</math>, and let <math>C_k</math> be the closure of the tail of this sequence.)
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