Closed subset of a metric space admits open subsets

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I have the following question and I require some help.



Let $(X,d)$ be a metric space. Let $C$ be a closed subset of $X$, and let $x in X$ be such that $x notin C$. Show that there exist disjoint subsets $U, V$ of $X$ with $ C subseteq U$ and $x in V$.



I am also given the following hint:

Hint: you may consider the function $f: X rightarrow mathbbR$ defined by
beginequation
f(y)=inf, d(y,c): c in C
endequation

I have managed to show that the function in the hint is continuous, which I think will help to use it in combination with the fact that the inverse image of an open set is open, but I cannot exactly move forward from that.



Any solutions or suggestions will be helpful.

Thanks.




general-topology metric-spaces




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  • 3




    One more hint: $f(x) >0$, so you can half it.
    – Berci
    Aug 8 at 18:27










  • Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
    – Dog_69
    Aug 8 at 20:05















up vote
0
down vote

favorite












I have the following question and I require some help.



Let $(X,d)$ be a metric space. Let $C$ be a closed subset of $X$, and let $x in X$ be such that $x notin C$. Show that there exist disjoint subsets $U, V$ of $X$ with $ C subseteq U$ and $x in V$.



I am also given the following hint:

Hint: you may consider the function $f: X rightarrow mathbbR$ defined by
beginequation
f(y)=inf, d(y,c): c in C
endequation

I have managed to show that the function in the hint is continuous, which I think will help to use it in combination with the fact that the inverse image of an open set is open, but I cannot exactly move forward from that.



Any solutions or suggestions will be helpful.

Thanks.




general-topology metric-spaces




share|cite|improve this question


















  • 3




    One more hint: $f(x) >0$, so you can half it.
    – Berci
    Aug 8 at 18:27










  • Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
    – Dog_69
    Aug 8 at 20:05













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following question and I require some help.



Let $(X,d)$ be a metric space. Let $C$ be a closed subset of $X$, and let $x in X$ be such that $x notin C$. Show that there exist disjoint subsets $U, V$ of $X$ with $ C subseteq U$ and $x in V$.



I am also given the following hint:

Hint: you may consider the function $f: X rightarrow mathbbR$ defined by
beginequation
f(y)=inf, d(y,c): c in C
endequation

I have managed to show that the function in the hint is continuous, which I think will help to use it in combination with the fact that the inverse image of an open set is open, but I cannot exactly move forward from that.



Any solutions or suggestions will be helpful.

Thanks.




general-topology metric-spaces




share|cite|improve this question














I have the following question and I require some help.



Let $(X,d)$ be a metric space. Let $C$ be a closed subset of $X$, and let $x in X$ be such that $x notin C$. Show that there exist disjoint subsets $U, V$ of $X$ with $ C subseteq U$ and $x in V$.



I am also given the following hint:

Hint: you may consider the function $f: X rightarrow mathbbR$ defined by
beginequation
f(y)=inf, d(y,c): c in C
endequation

I have managed to show that the function in the hint is continuous, which I think will help to use it in combination with the fact that the inverse image of an open set is open, but I cannot exactly move forward from that.



Any solutions or suggestions will be helpful.

Thanks.





general-topology metric-spaces





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 8 at 18:31









Clayton

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asked Aug 8 at 18:23









Nick

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444







  • 3




    One more hint: $f(x) >0$, so you can half it.
    – Berci
    Aug 8 at 18:27










  • Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
    – Dog_69
    Aug 8 at 20:05













  • 3




    One more hint: $f(x) >0$, so you can half it.
    – Berci
    Aug 8 at 18:27










  • Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
    – Dog_69
    Aug 8 at 20:05








3




3




One more hint: $f(x) >0$, so you can half it.
– Berci
Aug 8 at 18:27




One more hint: $f(x) >0$, so you can half it.
– Berci
Aug 8 at 18:27












Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
– Dog_69
Aug 8 at 20:05





Why do you not take $U=C$ and $V=Xsetminus C$? PD: the answer is the same even if $C$ would be open.
– Dog_69
Aug 8 at 20:05











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Take $f^-1(fracf(y)2,infty)$ for $V$ ; and let $U=Xsetminus overlineV$.



(You could also just take $V=B_y(fracf(y)2)$, the open ball of radius $fracf(y)2$ centered at $y$...)






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    up vote
    0
    down vote













    Take $f^-1(fracf(y)2,infty)$ for $V$ ; and let $U=Xsetminus overlineV$.



    (You could also just take $V=B_y(fracf(y)2)$, the open ball of radius $fracf(y)2$ centered at $y$...)






    share|cite|improve this answer


























      up vote
      0
      down vote













      Take $f^-1(fracf(y)2,infty)$ for $V$ ; and let $U=Xsetminus overlineV$.



      (You could also just take $V=B_y(fracf(y)2)$, the open ball of radius $fracf(y)2$ centered at $y$...)






      share|cite|improve this answer
























        up vote
        0
        down vote










        up vote
        0
        down vote









        Take $f^-1(fracf(y)2,infty)$ for $V$ ; and let $U=Xsetminus overlineV$.



        (You could also just take $V=B_y(fracf(y)2)$, the open ball of radius $fracf(y)2$ centered at $y$...)






        share|cite|improve this answer














        Take $f^-1(fracf(y)2,infty)$ for $V$ ; and let $U=Xsetminus overlineV$.



        (You could also just take $V=B_y(fracf(y)2)$, the open ball of radius $fracf(y)2$ centered at $y$...)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 8 at 21:18

























        answered Aug 8 at 19:13









        Chris Custer

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