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A proof goes as follows. Since the diameters tend to zero, the diameter of the intersection of the ''C''<sub>''n''</sub> 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 ''x''<sub>''n''</sub> of ''C''<sub>''n''</sub> for each ''n''. Since the diameter of ''C''<sub>''n''</sub> tends to zero and the ''C''<sub>''n''</sub> are nested, the ''x''<sub>''n''</sub> form a Cauchy sequence. Since the metric space is complete this Cauchy sequence converges to some point ''x''. Since each ''C''<sub>''n''</sub> is closed, and ''x'' is a limit of a sequence in ''C''<sub>''n''</sub>, ''x'' must lie in ''C''<sub>''n''</sub>. This is true for every ''n'', and therefore the intersection of the ''C''<sub>''n''</sub> must contain ''x''.
A converse to this theorem is also true: if ''X'' is a metric space with the property that the intersection of any nested family of nonempty closed subsets whose diameters tend to zero is non-empty, then ''X'' is a complete metric space. (To prove this, let ''x''<sub>''n''</sub> be a Cauchy sequence in ''X'', and let ''C''<sub>''n''</sub> be the closure of the tail of this sequence.)
== References ==
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