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Show that any Cauchy sequence in a metric space is always bounded.


Let \{x_{j}\}_{{j=1}}^{\infty }\, be a Cauchy sequence in the metric space (X,d)\,.


Since \{x_{j}\}\, is Cauchy, there exists an M\, such that d(x_{{M+1}},x_{j})<1\, for all j>M\,. For 1\leq j\leq M\,, let \delta _{j}=d(x_{{M+1}},x_{j})\,.


Now choose \delta >\max _{{j\in (1,...,M)}}\{1,\delta _{j}\}\,. Let B=B(x_{{M+1}},\delta )\subset X\,. Notice for 1\leq j\leq M,d(x_{{M+1}},x_{j})=\delta _{j}<\delta \,, so x_{j}\in B\,. For j>M,d(x_{{M+1}},x_{j})<1<\delta \,, so x_{j}\in B\,.


We have \{x_{j}\}\subset B\, making \{x_{j}\}\, bounded.


In general, any Cauchy sequence in a metric space is bounded.


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