Let $Y$ be a complete metric space. Then $C^0 (X,Y)$ is complete under the uniform convergence metric.












2












$begingroup$


I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.



Just to be clear, the metric is:



$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$



And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.



So far what I got is:



$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?



Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.



Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:



$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$



$blacksquare$



We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.



And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.






real-analysis metric-spaces uniform-convergence







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This question has an open bounty worth +50
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  • $begingroup$
    Feel free to ask questions on my answer
    $endgroup$
    – qbert
    yesterday
















2












$begingroup$


I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.



Just to be clear, the metric is:



$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$



And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.



So far what I got is:



$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?



Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.



Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:



$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$



$blacksquare$



We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.



And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.






real-analysis metric-spaces uniform-convergence







share|cite|improve this question











$endgroup$





This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.


The current answers do not contain enough detail.
















  • $begingroup$
    Feel free to ask questions on my answer
    $endgroup$
    – qbert
    yesterday














2












2








2


1



$begingroup$


I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.



Just to be clear, the metric is:



$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$



And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.



So far what I got is:



$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?



Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.



Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:



$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$



$blacksquare$



We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.



And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.






real-analysis metric-spaces uniform-convergence







share|cite|improve this question











$endgroup$




I'm trying to show that the set of continuous functions $f: X to Y$ is complete under the uniform convergence metric if $Y$ is complete.



Just to be clear, the metric is:



$$d(f,g) = text{sup}, {e(f(x), g(x)), ; ,xin X } $$



And $e(f(x), g(x))$ is the euclidian distance between $f(x)$ and $g(x)$.



So far what I got is:



$C^0(X,Y)$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?



Lemma: Let $f_n in C^0:(X,Y)$, $f:X to Y$, $underset{n to infty} lim d_n(f_n, f) = 0$, then $f$ is continuous.



Proof: Take $x_0 in X$ and $epsilon > 0$. There exists $n_0 in mathbb{N},; , n > n_0 implies d_n(f_n, f) < frac{epsilon}{5}$. We know $f_{n_0}$ is continuous. Then there exists $delta > 0$ such that the euclidian distance between $f_{n_0}(x_0)$ and $f_{n_0}(x)$ is smaller than $frac{epsilon}{5}$. If the euclidian distance between $x$ and $x_0$ is smaller than $delta$, then we have:



$$d(f(x), f(x_0)) leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) leq frac{3}{5}epsilon < epsilon$$



$blacksquare$



We have then that any Cauchy sequence in $C^0(X,Y)$ converges - because the limit of a sequence of continuous functions with domain $X$ and range $Y$ is itself a continuous function, which makes $C^0(X,Y)$ complete.



And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $Y$ is complete, for instance. $Y$ being complete guarantees that any function $f in C^0(X,Y)$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.






real-analysis metric-spaces uniform-convergence




real-analysis metric-spaces uniform-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 5 '18 at 4:44







Pedro Cavalcante Oliveira

















asked Dec 4 '18 at 19:39









Pedro Cavalcante OliveiraPedro Cavalcante Oliveira

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This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.


The current answers do not contain enough detail.








This question has an open bounty worth +50
reputation from Pedro Cavalcante Oliveira ending in 5 days.


The current answers do not contain enough detail.














  • $begingroup$
    Feel free to ask questions on my answer
    $endgroup$
    – qbert
    yesterday


















  • $begingroup$
    Feel free to ask questions on my answer
    $endgroup$
    – qbert
    yesterday
















$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday




$begingroup$
Feel free to ask questions on my answer
$endgroup$
– qbert
yesterday










1 Answer
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oldest

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3












$begingroup$

You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.



You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.



To do this, use completeness of $Y$ and the uniform estimate
$$
sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
$$
for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.






share|cite|improve this answer









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    1 Answer
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    active

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    $begingroup$

    You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.



    You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.



    To do this, use completeness of $Y$ and the uniform estimate
    $$
    sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
    $$
    for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.



      You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.



      To do this, use completeness of $Y$ and the uniform estimate
      $$
      sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
      $$
      for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.



        You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.



        To do this, use completeness of $Y$ and the uniform estimate
        $$
        sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
        $$
        for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.






        share|cite|improve this answer









        $endgroup$



        You are right to be skeptical. One tip off is you haven't used the completeness of $Y$ anywhere.



        You have shown that if a sequence $f_nto f$ with this metric, then $f$ is continuous. This is not the same as proving that if $f_n$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.



        To do this, use completeness of $Y$ and the uniform estimate
        $$
        sup_{xin X}|f_{n}(x)-f_m(x)|<epsilon
        $$
        for $n,mgeq N$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 19:46









        qbertqbert

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