How to prove a set of solution of a continuous function is closed?

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2
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E.g. if the set is $S=(x,y), g(x,y) = 0$, $g$ is a continuous function. How to prove that S is closed?



It seems that the relationship between continuous function and convergent sequence and the Lemma that "A set A in a metric space is closed iff the limit of every convergent
sequence in A belongs to A" can be used here. I am not sure how the relationship between continuous function and convergent sequence is applicable here.






general-topology continuity







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  • 1




    Possible duplicate of Prove: the set of zeros of a continuous function is closed.
    – Martin R
    2 hours ago










  • As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
    – Maik Pickl
    1 hour ago














up vote
2
down vote

favorite












E.g. if the set is $S=(x,y), g(x,y) = 0$, $g$ is a continuous function. How to prove that S is closed?



It seems that the relationship between continuous function and convergent sequence and the Lemma that "A set A in a metric space is closed iff the limit of every convergent
sequence in A belongs to A" can be used here. I am not sure how the relationship between continuous function and convergent sequence is applicable here.






general-topology continuity







share|cite|improve this question



















  • 1




    Possible duplicate of Prove: the set of zeros of a continuous function is closed.
    – Martin R
    2 hours ago










  • As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
    – Maik Pickl
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











E.g. if the set is $S=(x,y), g(x,y) = 0$, $g$ is a continuous function. How to prove that S is closed?



It seems that the relationship between continuous function and convergent sequence and the Lemma that "A set A in a metric space is closed iff the limit of every convergent
sequence in A belongs to A" can be used here. I am not sure how the relationship between continuous function and convergent sequence is applicable here.






general-topology continuity







share|cite|improve this question















E.g. if the set is $S=(x,y), g(x,y) = 0$, $g$ is a continuous function. How to prove that S is closed?



It seems that the relationship between continuous function and convergent sequence and the Lemma that "A set A in a metric space is closed iff the limit of every convergent
sequence in A belongs to A" can be used here. I am not sure how the relationship between continuous function and convergent sequence is applicable here.






general-topology continuity




general-topology continuity






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 1 hour ago









Andrés E. Caicedo

63.6k7156238




63.6k7156238










asked 2 hours ago









Aqqqq

1497




1497







  • 1




    Possible duplicate of Prove: the set of zeros of a continuous function is closed.
    – Martin R
    2 hours ago










  • As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
    – Maik Pickl
    1 hour ago












  • 1




    Possible duplicate of Prove: the set of zeros of a continuous function is closed.
    – Martin R
    2 hours ago










  • As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
    – Maik Pickl
    1 hour ago







1




1




Possible duplicate of Prove: the set of zeros of a continuous function is closed.
– Martin R
2 hours ago




Possible duplicate of Prove: the set of zeros of a continuous function is closed.
– Martin R
2 hours ago












As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
– Maik Pickl
1 hour ago




As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it.
– Maik Pickl
1 hour ago










3 Answers
3






active

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up vote
3
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Suppose $X$ is a general metric space, let $g:XrightarrowmathbbR$ be a real continuous function and let $A=g^-1(0)$. Choose a convergent sequence $(x_n)_ninmathbbN$ in $A$. Then by continuity
$$
gleft(lim_nrightarrowinftyx_nright) = lim_nrightarrowinftygleft(x_nright).
$$
Can you finish?






share|cite|improve this answer




















  • How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
    – Aqqqq
    1 hour ago

















up vote
2
down vote













Or you could just note that $0$ is a closed set. Hence $S=g^-1(0)$ is closed.






share|cite|improve this answer



























    up vote
    1
    down vote













    Take $(x,y)in S^c$, then $g(x,y)neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^-1[(b,c)]$ must be open. And we can easly see that $(x,y)in g^-1[(b,c)] subset S^c$. So $S^c$ is open then $S$ is closed.






    share|cite|improve this answer




















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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      up vote
      3
      down vote













      Suppose $X$ is a general metric space, let $g:XrightarrowmathbbR$ be a real continuous function and let $A=g^-1(0)$. Choose a convergent sequence $(x_n)_ninmathbbN$ in $A$. Then by continuity
      $$
      gleft(lim_nrightarrowinftyx_nright) = lim_nrightarrowinftygleft(x_nright).
      $$
      Can you finish?






      share|cite|improve this answer




















      • How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
        – Aqqqq
        1 hour ago














      up vote
      3
      down vote













      Suppose $X$ is a general metric space, let $g:XrightarrowmathbbR$ be a real continuous function and let $A=g^-1(0)$. Choose a convergent sequence $(x_n)_ninmathbbN$ in $A$. Then by continuity
      $$
      gleft(lim_nrightarrowinftyx_nright) = lim_nrightarrowinftygleft(x_nright).
      $$
      Can you finish?






      share|cite|improve this answer




















      • How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
        – Aqqqq
        1 hour ago












      up vote
      3
      down vote










      up vote
      3
      down vote









      Suppose $X$ is a general metric space, let $g:XrightarrowmathbbR$ be a real continuous function and let $A=g^-1(0)$. Choose a convergent sequence $(x_n)_ninmathbbN$ in $A$. Then by continuity
      $$
      gleft(lim_nrightarrowinftyx_nright) = lim_nrightarrowinftygleft(x_nright).
      $$
      Can you finish?






      share|cite|improve this answer












      Suppose $X$ is a general metric space, let $g:XrightarrowmathbbR$ be a real continuous function and let $A=g^-1(0)$. Choose a convergent sequence $(x_n)_ninmathbbN$ in $A$. Then by continuity
      $$
      gleft(lim_nrightarrowinftyx_nright) = lim_nrightarrowinftygleft(x_nright).
      $$
      Can you finish?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 2 hours ago









      Marc

      5,6071922




      5,6071922











      • How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
        – Aqqqq
        1 hour ago
















      • How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
        – Aqqqq
        1 hour ago















      How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
      – Aqqqq
      1 hour ago




      How do you know that a convergent sequence $(x_n)_ninmathbbN$ exist in $A$?
      – Aqqqq
      1 hour ago










      up vote
      2
      down vote













      Or you could just note that $0$ is a closed set. Hence $S=g^-1(0)$ is closed.






      share|cite|improve this answer
























        up vote
        2
        down vote













        Or you could just note that $0$ is a closed set. Hence $S=g^-1(0)$ is closed.






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          Or you could just note that $0$ is a closed set. Hence $S=g^-1(0)$ is closed.






          share|cite|improve this answer












          Or you could just note that $0$ is a closed set. Hence $S=g^-1(0)$ is closed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Chris Custer

          7,2622623




          7,2622623




















              up vote
              1
              down vote













              Take $(x,y)in S^c$, then $g(x,y)neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^-1[(b,c)]$ must be open. And we can easly see that $(x,y)in g^-1[(b,c)] subset S^c$. So $S^c$ is open then $S$ is closed.






              share|cite|improve this answer
























                up vote
                1
                down vote













                Take $(x,y)in S^c$, then $g(x,y)neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^-1[(b,c)]$ must be open. And we can easly see that $(x,y)in g^-1[(b,c)] subset S^c$. So $S^c$ is open then $S$ is closed.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Take $(x,y)in S^c$, then $g(x,y)neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^-1[(b,c)]$ must be open. And we can easly see that $(x,y)in g^-1[(b,c)] subset S^c$. So $S^c$ is open then $S$ is closed.






                  share|cite|improve this answer












                  Take $(x,y)in S^c$, then $g(x,y)neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^-1[(b,c)]$ must be open. And we can easly see that $(x,y)in g^-1[(b,c)] subset S^c$. So $S^c$ is open then $S$ is closed.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  S.S.Danyal

                  3113




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