Use the Lebesgue number lemma to prove that $f$ is uniformly continuous if $M$ is compact and $f$ continuous
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Suppose that $M$ is covering compact and that $f: M to N$ is continuous. Use the Lebesgue number lemma to prove that $f$ is uniformly continuous. Given $epsilon >0$ , the set $mathcal{U} = {B_{epsilon/2}(p) mid p in N}$ is an open covering of $f(M)$ . Since $f$ is continuous, $f^{-1}(B_{epsilon/2}(p))$ is open. But $M$ is compact, then $f(M)$ too. Thus, we can find a finite subcovering $$f(M) subset B_{epsilon/2}(p_{1})cup cdots cup B_{epsilon/2}(p_{n}).$$ Therefore, $$f^{-1}(B_{epsilon/2}(p_{1}))cup cdots cup f^{-1}(B_{epsilon/2}(p_{n}))$$ is a finite subcovering of $M$ . By the Lebesgue nummber lema, there is a $delta>0$ such that for every $x in B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$ for some $i$ . So, if $d(x,y) < delta$ , the " $y$ " are in some $B_{delta}(x) subse