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 $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$



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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 $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$



    Is this correct?










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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 $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$



      Is this correct?










      share|cite|improve this question
















      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 $B_{delta}(x) subset f^{-1}(B_{epsilon/2}(p_{i}))$, then $d(f(x),f(y)) leq d(f(x),p) + d(f(y),p) < epsilon.$



      Is this correct?







      proof-verification continuity metric-spaces compactness






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      edited Dec 28 '18 at 20:06









      Hayk

      2,1171213




      2,1171213










      asked Dec 28 '18 at 19:33









      Lucas CorrêaLucas Corrêa

      1,5521321




      1,5521321






















          1 Answer
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          The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.



          So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.



          If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.






          share|cite|improve this answer





















          • Thank you! Your answer is much clearer
            – Lucas Corrêa
            Dec 29 '18 at 2:05











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.



          So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.



          If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.






          share|cite|improve this answer





















          • Thank you! Your answer is much clearer
            – Lucas Corrêa
            Dec 29 '18 at 2:05
















          2














          The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.



          So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.



          If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.






          share|cite|improve this answer





















          • Thank you! Your answer is much clearer
            – Lucas Corrêa
            Dec 29 '18 at 2:05














          2












          2








          2






          The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.



          So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.



          If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.






          share|cite|improve this answer












          The Lebesgue number lemma I know (e.g. as stated here) needs no finite cover but any open cover.



          So just use the open cover ${f^{-1}[B(f(x), frac{varepsilon}{2})]: x in M}$ for $M$ and find $delta >0$ such that any set with diameter $< delta$ sits inside one member of the cover.



          If then $d(x,x') < delta$ apply this to ${x,x'}$, so that for some $p in M$ we have ${x,x'} subseteq f^{-1}[B(f(p), frac{varepsilon}{2})]$ which means $f(x), f(x') in B(f(p), frac{varepsilon}{2})$ and then the triangle inequality via $f(p)$ yields $d(f(x), f(x')) < varepsilon$, as required.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 22:55









          Henno BrandsmaHenno Brandsma

          105k347114




          105k347114












          • Thank you! Your answer is much clearer
            – Lucas Corrêa
            Dec 29 '18 at 2:05


















          • Thank you! Your answer is much clearer
            – Lucas Corrêa
            Dec 29 '18 at 2:05
















          Thank you! Your answer is much clearer
          – Lucas Corrêa
          Dec 29 '18 at 2:05




          Thank you! Your answer is much clearer
          – Lucas Corrêa
          Dec 29 '18 at 2:05


















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