$X$ compact metric implies $X$ separable












1












$begingroup$


I want to show that a compact metric space is a separable space.



Attempt:
Suppose $X$ is compact. Then it is complete and totally bounded. Taking $epsilon=frac{1}{n}$ in the definition of total boundedness gives me $exists x_1,x_2,ldots,$ such that $Xsubset bigcup_{i=1}^nB_{frac{1}{n}}(x_i)$..



how to continue?






real-analysis general-topology metric-spaces compactness separable-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
    $endgroup$
    – Xiao
    Dec 1 '14 at 16:57












  • $begingroup$
    I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
    $endgroup$
    – Jon
    Dec 1 '14 at 17:03










  • $begingroup$
    Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
    $endgroup$
    – Xiao
    Dec 1 '14 at 17:43












  • $begingroup$
    @Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
    $endgroup$
    – Luiz Cordeiro
    Dec 1 '14 at 19:01










  • $begingroup$
    Compact does not imply separable, you may consider inserting the word metric in the title of your question.
    $endgroup$
    – Mirko
    Dec 1 '14 at 20:24
















1












$begingroup$


I want to show that a compact metric space is a separable space.



Attempt:
Suppose $X$ is compact. Then it is complete and totally bounded. Taking $epsilon=frac{1}{n}$ in the definition of total boundedness gives me $exists x_1,x_2,ldots,$ such that $Xsubset bigcup_{i=1}^nB_{frac{1}{n}}(x_i)$..



how to continue?






real-analysis general-topology metric-spaces compactness separable-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
    $endgroup$
    – Xiao
    Dec 1 '14 at 16:57












  • $begingroup$
    I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
    $endgroup$
    – Jon
    Dec 1 '14 at 17:03










  • $begingroup$
    Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
    $endgroup$
    – Xiao
    Dec 1 '14 at 17:43












  • $begingroup$
    @Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
    $endgroup$
    – Luiz Cordeiro
    Dec 1 '14 at 19:01










  • $begingroup$
    Compact does not imply separable, you may consider inserting the word metric in the title of your question.
    $endgroup$
    – Mirko
    Dec 1 '14 at 20:24














1












1








1





$begingroup$


I want to show that a compact metric space is a separable space.



Attempt:
Suppose $X$ is compact. Then it is complete and totally bounded. Taking $epsilon=frac{1}{n}$ in the definition of total boundedness gives me $exists x_1,x_2,ldots,$ such that $Xsubset bigcup_{i=1}^nB_{frac{1}{n}}(x_i)$..



how to continue?






real-analysis general-topology metric-spaces compactness separable-spaces







share|cite|improve this question











$endgroup$




I want to show that a compact metric space is a separable space.



Attempt:
Suppose $X$ is compact. Then it is complete and totally bounded. Taking $epsilon=frac{1}{n}$ in the definition of total boundedness gives me $exists x_1,x_2,ldots,$ such that $Xsubset bigcup_{i=1}^nB_{frac{1}{n}}(x_i)$..



how to continue?






real-analysis general-topology metric-spaces compactness separable-spaces




real-analysis general-topology metric-spaces compactness separable-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 12:13









Davide Giraudo

125k16150261




125k16150261










asked Dec 1 '14 at 16:53









JonJon

61




61








  • 2




    $begingroup$
    Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
    $endgroup$
    – Xiao
    Dec 1 '14 at 16:57












  • $begingroup$
    I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
    $endgroup$
    – Jon
    Dec 1 '14 at 17:03










  • $begingroup$
    Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
    $endgroup$
    – Xiao
    Dec 1 '14 at 17:43












  • $begingroup$
    @Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
    $endgroup$
    – Luiz Cordeiro
    Dec 1 '14 at 19:01










  • $begingroup$
    Compact does not imply separable, you may consider inserting the word metric in the title of your question.
    $endgroup$
    – Mirko
    Dec 1 '14 at 20:24














  • 2




    $begingroup$
    Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
    $endgroup$
    – Xiao
    Dec 1 '14 at 16:57












  • $begingroup$
    I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
    $endgroup$
    – Jon
    Dec 1 '14 at 17:03










  • $begingroup$
    Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
    $endgroup$
    – Xiao
    Dec 1 '14 at 17:43












  • $begingroup$
    @Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
    $endgroup$
    – Luiz Cordeiro
    Dec 1 '14 at 19:01










  • $begingroup$
    Compact does not imply separable, you may consider inserting the word metric in the title of your question.
    $endgroup$
    – Mirko
    Dec 1 '14 at 20:24








2




2




$begingroup$
Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
$endgroup$
– Xiao
Dec 1 '14 at 16:57






$begingroup$
Let $A_n:= {x_1,...x_n}$, the centers of the finite subcovers for balls with radius $1/n$. Then take $cup_{nin mathbb{N}} A_n$, this set is countable and dense.
$endgroup$
– Xiao
Dec 1 '14 at 16:57














$begingroup$
I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
$endgroup$
– Jon
Dec 1 '14 at 17:03




$begingroup$
I understand why it is countable since the union of countable many points is once again countable. How do we know that set is dense though?
$endgroup$
– Jon
Dec 1 '14 at 17:03












$begingroup$
Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
$endgroup$
– Xiao
Dec 1 '14 at 17:43






$begingroup$
Try to show that for each $epsilon > 0$ and $xin X$, there exists $x_epsilon in X$ such that $d(x, x_epsilon) < epsilon$. Hint: there is a cover of $X$ with balls of radius $1/n < epsilon$, also that the center of these balls are in $A = cup_n A_n$.
$endgroup$
– Xiao
Dec 1 '14 at 17:43














$begingroup$
@Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
$endgroup$
– Luiz Cordeiro
Dec 1 '14 at 19:01




$begingroup$
@Xiao Actually $A_n=left{x_1,ldots,x_{N_n}right}$, to avoid confusions.
$endgroup$
– Luiz Cordeiro
Dec 1 '14 at 19:01












$begingroup$
Compact does not imply separable, you may consider inserting the word metric in the title of your question.
$endgroup$
– Mirko
Dec 1 '14 at 20:24




$begingroup$
Compact does not imply separable, you may consider inserting the word metric in the title of your question.
$endgroup$
– Mirko
Dec 1 '14 at 20:24










1 Answer
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$begingroup$

If $X$ is compact, there are points ${x_i^{(n)}}_{i=1}^{k_n}$ for all $n in mathbb N$ and finite $k_n$, such that
$$
X subset bigcup_{i=1}^{k_n} B(x_i,1/n) qquad (n in mathbb N)
$$
Now take the union of all the $x_i^{(n)}$ to get a countable dense subset of $X$.






share|cite|improve this answer









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    $begingroup$

    If $X$ is compact, there are points ${x_i^{(n)}}_{i=1}^{k_n}$ for all $n in mathbb N$ and finite $k_n$, such that
    $$
    X subset bigcup_{i=1}^{k_n} B(x_i,1/n) qquad (n in mathbb N)
    $$
    Now take the union of all the $x_i^{(n)}$ to get a countable dense subset of $X$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $X$ is compact, there are points ${x_i^{(n)}}_{i=1}^{k_n}$ for all $n in mathbb N$ and finite $k_n$, such that
      $$
      X subset bigcup_{i=1}^{k_n} B(x_i,1/n) qquad (n in mathbb N)
      $$
      Now take the union of all the $x_i^{(n)}$ to get a countable dense subset of $X$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $X$ is compact, there are points ${x_i^{(n)}}_{i=1}^{k_n}$ for all $n in mathbb N$ and finite $k_n$, such that
        $$
        X subset bigcup_{i=1}^{k_n} B(x_i,1/n) qquad (n in mathbb N)
        $$
        Now take the union of all the $x_i^{(n)}$ to get a countable dense subset of $X$.






        share|cite|improve this answer









        $endgroup$



        If $X$ is compact, there are points ${x_i^{(n)}}_{i=1}^{k_n}$ for all $n in mathbb N$ and finite $k_n$, such that
        $$
        X subset bigcup_{i=1}^{k_n} B(x_i,1/n) qquad (n in mathbb N)
        $$
        Now take the union of all the $x_i^{(n)}$ to get a countable dense subset of $X$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '14 at 21:01









        EpsilonEpsilon

        3,50332238




        3,50332238






























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