# 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 $d(f(x_{k}),f(y_{k}))\geq\epsilon$.

Title proof of Heine-Cantor theorem ProofOfHeineCantorTheorem 2013-03-22 13:31:26 2013-03-22 13:31:26 paolini (1187) paolini (1187) 5 paolini (1187) Proof msc 46A99