proof of Dini’s theorem

Without loss of generality we will assume that $X$ is compact and, by replacing $f_{n}$ with $f-f_{n}$ , that the net converges monotonically to 0.

Let $\\epsilon>0$ .For each $x\\in X$ , we can choose an $n_{x}$ , such that $f_{n_{x}}(x)<\\epsilon/2$ . Since $f_{n_{x}}$ is continuous,there is an openneighbourhood $U_{x}$ of $x$ , such that for each $y\\in U_{x}$ , we have $f_{n_{x}}(y)<\\epsilon/2$ . The open sets $U_{x}$ cover $X$ , which is compact, so we can choosefinitely many $x_{1},\\ldots,x_{k}$ such that the $U_{x_{i}}$ also cover $X$ . Then,if $N\\geq n_{x_{1}},\\ldots,n_{x_{k}}$ , we have $f_{n}(x)<\\epsilon$ for each $n\\geq N$ and $x\\in X$ , since the sequence $f_{n}$ is monotonically decreasing.Thus, $\\{f_{n}\\}$ converges to 0 uniformly on $X$ , which was to be proven.

单词	ProofOfDinisTheorem
释义	proof of Dini’s theorem Without loss of generality we will assume that $X$ is compact and, by replacing $f_{n}$ with $f-f_{n}$ , that the net converges monotonically to 0. Let $\\epsilon>0$ .For each $x\\in X$ , we can choose an $n_{x}$ , such that $f_{n_{x}}(x)<\\epsilon/2$ . Since $f_{n_{x}}$ is continuous,there is an openneighbourhood $U_{x}$ of $x$ , such that for each $y\\in U_{x}$ , we have $f_{n_{x}}(y)<\\epsilon/2$ . The open sets $U_{x}$ cover $X$ , which is compact, so we can choosefinitely many $x_{1},\\ldots,x_{k}$ such that the $U_{x_{i}}$ also cover $X$ . Then,if $N\\geq n_{x_{1}},\\ldots,n_{x_{k}}$ , we have $f_{n}(x)<\\epsilon$ for each $n\\geq N$ and $x\\in X$ , since the sequence $f_{n}$ is monotonically decreasing.Thus, $\\{f_{n}\\}$ converges to 0 uniformly on $X$ , which was to be proven.
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