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Assume \sup \left\{|x_{n}(t)-x(t)|:t\in [0,1]\right\}\to 0\, as n\to \infty \,. Prove that x(t)\, is continuous when x_{n}(t)\, is continuous for all n\,.


Suppose x_{n}(t)\, is continuous for all t\in [0,1]\,, and all n\,. Let t_{0}\in [0,1]\,. Then |x_{0}(t)-x_{n}(t_{0})|\to 0\, as t\to t_{0}\, for all n\,. Let \epsilon >0\,. There exists an N\, such that whenever n>N\,, \sup _{{t\in [0,1]}}|x_{n}(t)-x(t)|<{\frac  {\epsilon }{3}}\,. Let n=N+1\, so that there is no dependence on n\,. There exists a \delta \, such that |x_{{N+1}}(t)-x_{{N+1}}(t_{0})|<{\frac  {\epsilon }{3}}\, whenever |t-t_{0}|<\delta ,t\in [0,1]\,.


Let |t-t_{0}|<\delta ,t\in [0,1],n>N\,. Then by the triangle inequality,

|x(t)-x(t_{0})|\leq |x(t)-x_{{N+1}}(t)|+|x_{{N+1}}(t)-x_{{N+1}}(t_{0})|+|x_{{N+1}}(t_{0})-x(t_{0})|\,

\leq {\frac  {\epsilon }{3}}+{\frac  {\epsilon }{3}}+{\frac  {\epsilon }{3}}=\epsilon \,


Since t_{0}\, was arbitrary, x\, is continuous whenever x_{n}\, is continuous for all n\,.


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