For continuous functions $f_1:Xto Y_1$ and $f_2:Xto Y_2$ prove $f_1nabla f_2:Xto Y_1times Y_2$ is continuous












1















For functions $f_1:Xto Y_1$ and $f_2:Xto Y_2$ define $f_1nabla f_2:Xto Y_1times Y_2$ by $(f_1nabla f_2)(x)=(f_1(x),f_2(x))$.



Prove that $f_1nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.




(Link to the image that replaced text).



So, $f_1nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.



Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.



Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?






analysis continuity







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  • How is the distance function defined in $Y_1times Y_2$?
    – Frpzzd
    Dec 4 '18 at 21:21










  • X, Y1, and Y2 are all topological spaces
    – general1597
    Dec 4 '18 at 21:27










  • "... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
    – zipirovich
    Dec 4 '18 at 22:37










  • Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
    – PtF
    Dec 4 '18 at 23:04


















1















For functions $f_1:Xto Y_1$ and $f_2:Xto Y_2$ define $f_1nabla f_2:Xto Y_1times Y_2$ by $(f_1nabla f_2)(x)=(f_1(x),f_2(x))$.



Prove that $f_1nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.




(Link to the image that replaced text).



So, $f_1nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.



Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.



Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?






analysis continuity







share|cite|improve this question
























  • How is the distance function defined in $Y_1times Y_2$?
    – Frpzzd
    Dec 4 '18 at 21:21










  • X, Y1, and Y2 are all topological spaces
    – general1597
    Dec 4 '18 at 21:27










  • "... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
    – zipirovich
    Dec 4 '18 at 22:37










  • Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
    – PtF
    Dec 4 '18 at 23:04
















1












1








1








For functions $f_1:Xto Y_1$ and $f_2:Xto Y_2$ define $f_1nabla f_2:Xto Y_1times Y_2$ by $(f_1nabla f_2)(x)=(f_1(x),f_2(x))$.



Prove that $f_1nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.




(Link to the image that replaced text).



So, $f_1nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.



Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.



Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?






analysis continuity







share|cite|improve this question
















For functions $f_1:Xto Y_1$ and $f_2:Xto Y_2$ define $f_1nabla f_2:Xto Y_1times Y_2$ by $(f_1nabla f_2)(x)=(f_1(x),f_2(x))$.



Prove that $f_1nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.




(Link to the image that replaced text).



So, $f_1nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.



Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.



Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?






analysis continuity




analysis continuity






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share|cite|improve this question













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edited Dec 4 '18 at 22:37









zipirovich

11.1k11631




11.1k11631










asked Dec 4 '18 at 21:19









general1597general1597

313




313












  • How is the distance function defined in $Y_1times Y_2$?
    – Frpzzd
    Dec 4 '18 at 21:21










  • X, Y1, and Y2 are all topological spaces
    – general1597
    Dec 4 '18 at 21:27










  • "... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
    – zipirovich
    Dec 4 '18 at 22:37










  • Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
    – PtF
    Dec 4 '18 at 23:04




















  • How is the distance function defined in $Y_1times Y_2$?
    – Frpzzd
    Dec 4 '18 at 21:21










  • X, Y1, and Y2 are all topological spaces
    – general1597
    Dec 4 '18 at 21:27










  • "... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
    – zipirovich
    Dec 4 '18 at 22:37










  • Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
    – PtF
    Dec 4 '18 at 23:04


















How is the distance function defined in $Y_1times Y_2$?
– Frpzzd
Dec 4 '18 at 21:21




How is the distance function defined in $Y_1times Y_2$?
– Frpzzd
Dec 4 '18 at 21:21












X, Y1, and Y2 are all topological spaces
– general1597
Dec 4 '18 at 21:27




X, Y1, and Y2 are all topological spaces
– general1597
Dec 4 '18 at 21:27












"... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
– zipirovich
Dec 4 '18 at 22:37




"... and each neighborhood $V$ of $f_1nabla f_2$ ... " doesn't make sense -- $f_1nabla f_2$ is a function, not one of the spaces.
– zipirovich
Dec 4 '18 at 22:37












Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
– PtF
Dec 4 '18 at 23:04






Hint: Given $Utimes Vsubset X_1times X_2$ then $(f_1nabla f_2)^{-1}(Utimes V)=f_1^{-1}(U)cap f_2^{-1}(V)$.
– PtF
Dec 4 '18 at 23:04












1 Answer
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You can prove this result:



If $pi_i: Y_1times Y_2 to Y_i$ is the projection on the $i$-th factor then a function
$g:Xto Y_1times Y_2$ is continuos if and only if $pi_1circ g$ and $pi_2circ g$ are continuos.



In your case you have that $pi_1circ (f_1nabla f_2)=f_1$ and $pi_2circ (f_1nabla f_2)=f_2 $ that are continuos so $f_1nabla f_2$ is continuos.



You can prove the results in this way:



If $g$ is continuos then $pi_1circ g$ and $pi_2circ g$ is continuos because $pi_1$ and $pi_2$ are continuos.



Conversely Suppose that $pi_1circ g$ and $pi_2circ g$ are continuos.
Let $A=Vtimes W$ a basic neighborhood of $Y_1times Y_2$.



Then $A=pi_1^{-1}(V)cap pi_2^{-1}(W)$ and so



$g^{-1}(A)=g^{-1}(pi_1^{-1}(V))cap g^{-1}(pi_2^{-1}(W))=$



$=(pi_1circ g)^{-1}(V)cap (pi_2circ g)^{-1}(W) $



That it is open






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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    0














    You can prove this result:



    If $pi_i: Y_1times Y_2 to Y_i$ is the projection on the $i$-th factor then a function
    $g:Xto Y_1times Y_2$ is continuos if and only if $pi_1circ g$ and $pi_2circ g$ are continuos.



    In your case you have that $pi_1circ (f_1nabla f_2)=f_1$ and $pi_2circ (f_1nabla f_2)=f_2 $ that are continuos so $f_1nabla f_2$ is continuos.



    You can prove the results in this way:



    If $g$ is continuos then $pi_1circ g$ and $pi_2circ g$ is continuos because $pi_1$ and $pi_2$ are continuos.



    Conversely Suppose that $pi_1circ g$ and $pi_2circ g$ are continuos.
    Let $A=Vtimes W$ a basic neighborhood of $Y_1times Y_2$.



    Then $A=pi_1^{-1}(V)cap pi_2^{-1}(W)$ and so



    $g^{-1}(A)=g^{-1}(pi_1^{-1}(V))cap g^{-1}(pi_2^{-1}(W))=$



    $=(pi_1circ g)^{-1}(V)cap (pi_2circ g)^{-1}(W) $



    That it is open






    share|cite|improve this answer




























      0














      You can prove this result:



      If $pi_i: Y_1times Y_2 to Y_i$ is the projection on the $i$-th factor then a function
      $g:Xto Y_1times Y_2$ is continuos if and only if $pi_1circ g$ and $pi_2circ g$ are continuos.



      In your case you have that $pi_1circ (f_1nabla f_2)=f_1$ and $pi_2circ (f_1nabla f_2)=f_2 $ that are continuos so $f_1nabla f_2$ is continuos.



      You can prove the results in this way:



      If $g$ is continuos then $pi_1circ g$ and $pi_2circ g$ is continuos because $pi_1$ and $pi_2$ are continuos.



      Conversely Suppose that $pi_1circ g$ and $pi_2circ g$ are continuos.
      Let $A=Vtimes W$ a basic neighborhood of $Y_1times Y_2$.



      Then $A=pi_1^{-1}(V)cap pi_2^{-1}(W)$ and so



      $g^{-1}(A)=g^{-1}(pi_1^{-1}(V))cap g^{-1}(pi_2^{-1}(W))=$



      $=(pi_1circ g)^{-1}(V)cap (pi_2circ g)^{-1}(W) $



      That it is open






      share|cite|improve this answer


























        0












        0








        0






        You can prove this result:



        If $pi_i: Y_1times Y_2 to Y_i$ is the projection on the $i$-th factor then a function
        $g:Xto Y_1times Y_2$ is continuos if and only if $pi_1circ g$ and $pi_2circ g$ are continuos.



        In your case you have that $pi_1circ (f_1nabla f_2)=f_1$ and $pi_2circ (f_1nabla f_2)=f_2 $ that are continuos so $f_1nabla f_2$ is continuos.



        You can prove the results in this way:



        If $g$ is continuos then $pi_1circ g$ and $pi_2circ g$ is continuos because $pi_1$ and $pi_2$ are continuos.



        Conversely Suppose that $pi_1circ g$ and $pi_2circ g$ are continuos.
        Let $A=Vtimes W$ a basic neighborhood of $Y_1times Y_2$.



        Then $A=pi_1^{-1}(V)cap pi_2^{-1}(W)$ and so



        $g^{-1}(A)=g^{-1}(pi_1^{-1}(V))cap g^{-1}(pi_2^{-1}(W))=$



        $=(pi_1circ g)^{-1}(V)cap (pi_2circ g)^{-1}(W) $



        That it is open






        share|cite|improve this answer














        You can prove this result:



        If $pi_i: Y_1times Y_2 to Y_i$ is the projection on the $i$-th factor then a function
        $g:Xto Y_1times Y_2$ is continuos if and only if $pi_1circ g$ and $pi_2circ g$ are continuos.



        In your case you have that $pi_1circ (f_1nabla f_2)=f_1$ and $pi_2circ (f_1nabla f_2)=f_2 $ that are continuos so $f_1nabla f_2$ is continuos.



        You can prove the results in this way:



        If $g$ is continuos then $pi_1circ g$ and $pi_2circ g$ is continuos because $pi_1$ and $pi_2$ are continuos.



        Conversely Suppose that $pi_1circ g$ and $pi_2circ g$ are continuos.
        Let $A=Vtimes W$ a basic neighborhood of $Y_1times Y_2$.



        Then $A=pi_1^{-1}(V)cap pi_2^{-1}(W)$ and so



        $g^{-1}(A)=g^{-1}(pi_1^{-1}(V))cap g^{-1}(pi_2^{-1}(W))=$



        $=(pi_1circ g)^{-1}(V)cap (pi_2circ g)^{-1}(W) $



        That it is open







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 22:53

























        answered Dec 4 '18 at 22:43









        Federico FalluccaFederico Fallucca

        1,81318




        1,81318






























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