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Homeproof of Heine-Cantor theorem

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# proof of Heine-Cantor theorem

We prove this theorem in the case when $X$ and $Y$ are metric spaces.

Suppose $f$ is not uniformly continuous. Then

$\exists\epsilon>0\ \forall\delta>0\ \exists x,y\in X\quad d(x,y)<\delta\ % \mathrm{but}\ d(f(x),f(y))\geq\epsilon.$ |

In particular by letting $\delta=1/k$ we can construct two sequences $x_{k}$ and $y_{k}$ such that

$d(x_{k},y_{k})<1/k\ \mathrm{and}\ d(f(x_{k}),f(y_{k})\geq\epsilon.$ |

Since $X$ is compact the two sequence have convergent subsequences i.e.

$x_{{k_{j}}}\to\bar{x}\in X,\quad y_{{k_{j}}}\to\bar{y}\in X.$ |

Since $d(x_{k},y_{k})\to 0$ we have $\bar{x}=\bar{y}$. Being $f$ continuous we hence conclude $d(f(x_{{k_{j}}}),f(y_{{k_{j}}}))\to 0$ which is a contradiction being $d(f(x_{k}),f(y_{k}))\geq\epsilon$.

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Proof

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## Mathematics Subject Classification

46A99*no label found*

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