Continuity at a point in terms of closure












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If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






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    If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



    I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



    Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






    general-topology continuity







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      $begingroup$


      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






      general-topology continuity







      share|cite|improve this question









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      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverline{A} implies f(x_0)inoverline{f(A)}$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^{-1}(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverline{A}$, we have $Acap f^{-1}(V)neqvarnothing$. Let $xin Acap f^{-1}(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverline{f(A)}$.






      general-topology continuity




      general-topology continuity






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      BlondCaféBlondCafé

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          Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






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            It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
            We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



            Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



            It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






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              Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






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                Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






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                  $begingroup$

                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






                  share|cite|improve this answer









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                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overline{f^{-1}(Ysetminus V)}$, then $f(x_o)in overline{f(f^{-1}(Ysetminus V))}subset overline{Ysetminus V}= Ysetminus V$, a contradiction, so $x_0 notin overline{f^{-1}(Ysetminus V)}$. Then, if $U = Xsetminus overline{f^{-1}(Ysetminus V)}$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.







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                  guchiheguchihe

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                      $begingroup$

                      It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                      We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                      Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                      It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                        We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                        Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                        It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






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                          2








                          2





                          $begingroup$

                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                          share|cite|improve this answer









                          $endgroup$



                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^{-1}[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overline{f^{-1}[Ysetminus V]}$ and so the assumption on $f$ would imply that $y=f(x_0) in overline{f[f^{-1}[Ysetminus V]]}$. But $f[f^{-1}[B]] subseteq B$ for any $B$ so we'd deduce that $y in overline{Ysetminus V} = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.







                          share|cite|improve this answer












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                          share|cite|improve this answer










                          answered yesterday









                          Henno BrandsmaHenno Brandsma

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