Is this definition a continuous map?












0












$begingroup$



Let (X, $mathscr{T_x}$) and (Y, $mathscr{T_y}$) be topological spaces. A map $f : X → Y$ is called continuous if the inverse image of each set open in $Y$ is open in X (that is $f^{−1}$ maps $mathscr{T_y}$ into $mathscr{T_x}$)




I want to understand what type of function is this if all this definition above



if the inverse image of each set open in Y is open in X



is also open in Y



In standard definition is continuous only if is open in X but it is not said that Y 'is opened' by continuity definition itself, i.e




{if the inverse image of each set open in Y is open in X} is opened in Y




In standard definition I see a closure, "the price to pay" is a 'closed Y' for a open X, not a opening Y from an opened X by Y




Y = the inverse image of each set open in $Y$ is open in X




In an equivalent way it is like saying that X is open by Y as inverse image of each set open in Y but this is a self-referential definition because is works like a closed cicle loop, only from Y to X, there is no return for both Y and X, if not exclusively for oneself: Y but not X



To have continuity the first subject is inverse image of each set open in Y (but Y can be 'closed') while X must be 'is open'.



This definition has only one verse, there is not a 'feedback' continuity in the "opposite" direction (which is not an opposite verse, but just an Y as open set respect to 'in Y is open in X')



I need a hetero-referential continuity (as a kind of a full duplex signal) not an self-referential continuity (as a half-duplex)



EDIT: open in Y is for me this










share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let (X, $mathscr{T_x}$) and (Y, $mathscr{T_y}$) be topological spaces. A map $f : X → Y$ is called continuous if the inverse image of each set open in $Y$ is open in X (that is $f^{−1}$ maps $mathscr{T_y}$ into $mathscr{T_x}$)




    I want to understand what type of function is this if all this definition above



    if the inverse image of each set open in Y is open in X



    is also open in Y



    In standard definition is continuous only if is open in X but it is not said that Y 'is opened' by continuity definition itself, i.e




    {if the inverse image of each set open in Y is open in X} is opened in Y




    In standard definition I see a closure, "the price to pay" is a 'closed Y' for a open X, not a opening Y from an opened X by Y




    Y = the inverse image of each set open in $Y$ is open in X




    In an equivalent way it is like saying that X is open by Y as inverse image of each set open in Y but this is a self-referential definition because is works like a closed cicle loop, only from Y to X, there is no return for both Y and X, if not exclusively for oneself: Y but not X



    To have continuity the first subject is inverse image of each set open in Y (but Y can be 'closed') while X must be 'is open'.



    This definition has only one verse, there is not a 'feedback' continuity in the "opposite" direction (which is not an opposite verse, but just an Y as open set respect to 'in Y is open in X')



    I need a hetero-referential continuity (as a kind of a full duplex signal) not an self-referential continuity (as a half-duplex)



    EDIT: open in Y is for me this










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let (X, $mathscr{T_x}$) and (Y, $mathscr{T_y}$) be topological spaces. A map $f : X → Y$ is called continuous if the inverse image of each set open in $Y$ is open in X (that is $f^{−1}$ maps $mathscr{T_y}$ into $mathscr{T_x}$)




      I want to understand what type of function is this if all this definition above



      if the inverse image of each set open in Y is open in X



      is also open in Y



      In standard definition is continuous only if is open in X but it is not said that Y 'is opened' by continuity definition itself, i.e




      {if the inverse image of each set open in Y is open in X} is opened in Y




      In standard definition I see a closure, "the price to pay" is a 'closed Y' for a open X, not a opening Y from an opened X by Y




      Y = the inverse image of each set open in $Y$ is open in X




      In an equivalent way it is like saying that X is open by Y as inverse image of each set open in Y but this is a self-referential definition because is works like a closed cicle loop, only from Y to X, there is no return for both Y and X, if not exclusively for oneself: Y but not X



      To have continuity the first subject is inverse image of each set open in Y (but Y can be 'closed') while X must be 'is open'.



      This definition has only one verse, there is not a 'feedback' continuity in the "opposite" direction (which is not an opposite verse, but just an Y as open set respect to 'in Y is open in X')



      I need a hetero-referential continuity (as a kind of a full duplex signal) not an self-referential continuity (as a half-duplex)



      EDIT: open in Y is for me this










      share|cite|improve this question











      $endgroup$





      Let (X, $mathscr{T_x}$) and (Y, $mathscr{T_y}$) be topological spaces. A map $f : X → Y$ is called continuous if the inverse image of each set open in $Y$ is open in X (that is $f^{−1}$ maps $mathscr{T_y}$ into $mathscr{T_x}$)




      I want to understand what type of function is this if all this definition above



      if the inverse image of each set open in Y is open in X



      is also open in Y



      In standard definition is continuous only if is open in X but it is not said that Y 'is opened' by continuity definition itself, i.e




      {if the inverse image of each set open in Y is open in X} is opened in Y




      In standard definition I see a closure, "the price to pay" is a 'closed Y' for a open X, not a opening Y from an opened X by Y




      Y = the inverse image of each set open in $Y$ is open in X




      In an equivalent way it is like saying that X is open by Y as inverse image of each set open in Y but this is a self-referential definition because is works like a closed cicle loop, only from Y to X, there is no return for both Y and X, if not exclusively for oneself: Y but not X



      To have continuity the first subject is inverse image of each set open in Y (but Y can be 'closed') while X must be 'is open'.



      This definition has only one verse, there is not a 'feedback' continuity in the "opposite" direction (which is not an opposite verse, but just an Y as open set respect to 'in Y is open in X')



      I need a hetero-referential continuity (as a kind of a full duplex signal) not an self-referential continuity (as a half-duplex)



      EDIT: open in Y is for me this







      continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 16:09







      Peter Long

















      asked Dec 16 '18 at 15:07









      Peter LongPeter Long

      194




      194






















          2 Answers
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          0












          $begingroup$

          I don't really understand what the issue is.
          We have topological spaces $(X,mathscr{T}_X)$ and $(Y,mathscr{T}_Y)$, and a function $f:X to Y$.



          $f$ is continuous iff



          $$forall O in mathscr{T}_Y: f^{-1}[O] in mathscr{T}_X$$



          where as usual $f^{-1}[O] = {x in X: f(x) in O}$, so this has nothing to do with an inverse function, just a so-called inverse image.



          So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11



















          0












          $begingroup$

          I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".



          You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11











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          2 Answers
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          2 Answers
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          active

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          0












          $begingroup$

          I don't really understand what the issue is.
          We have topological spaces $(X,mathscr{T}_X)$ and $(Y,mathscr{T}_Y)$, and a function $f:X to Y$.



          $f$ is continuous iff



          $$forall O in mathscr{T}_Y: f^{-1}[O] in mathscr{T}_X$$



          where as usual $f^{-1}[O] = {x in X: f(x) in O}$, so this has nothing to do with an inverse function, just a so-called inverse image.



          So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11
















          0












          $begingroup$

          I don't really understand what the issue is.
          We have topological spaces $(X,mathscr{T}_X)$ and $(Y,mathscr{T}_Y)$, and a function $f:X to Y$.



          $f$ is continuous iff



          $$forall O in mathscr{T}_Y: f^{-1}[O] in mathscr{T}_X$$



          where as usual $f^{-1}[O] = {x in X: f(x) in O}$, so this has nothing to do with an inverse function, just a so-called inverse image.



          So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11














          0












          0








          0





          $begingroup$

          I don't really understand what the issue is.
          We have topological spaces $(X,mathscr{T}_X)$ and $(Y,mathscr{T}_Y)$, and a function $f:X to Y$.



          $f$ is continuous iff



          $$forall O in mathscr{T}_Y: f^{-1}[O] in mathscr{T}_X$$



          where as usual $f^{-1}[O] = {x in X: f(x) in O}$, so this has nothing to do with an inverse function, just a so-called inverse image.



          So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.






          share|cite|improve this answer









          $endgroup$



          I don't really understand what the issue is.
          We have topological spaces $(X,mathscr{T}_X)$ and $(Y,mathscr{T}_Y)$, and a function $f:X to Y$.



          $f$ is continuous iff



          $$forall O in mathscr{T}_Y: f^{-1}[O] in mathscr{T}_X$$



          where as usual $f^{-1}[O] = {x in X: f(x) in O}$, so this has nothing to do with an inverse function, just a so-called inverse image.



          So we demand exactly that whenever we have an open set of $Y$ (i.e. an element of $mathscr{T}_Y$) its inverse image is open in $X$, i.e. an element of $mathscr{T}_X$. It does not say that $O$ is only open if that inverse image is. This is the definition of a stronger notion called "quotient map". Nothing is "opened", the set of open sets is just a given on both sides.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 15:19









          Henno BrandsmaHenno Brandsma

          110k348118




          110k348118












          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11


















          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11
















          $begingroup$
          open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
          $endgroup$
          – Peter Long
          Dec 16 '18 at 16:11




          $begingroup$
          open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
          $endgroup$
          – Peter Long
          Dec 16 '18 at 16:11











          0












          $begingroup$

          I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".



          You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11
















          0












          $begingroup$

          I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".



          You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11














          0












          0








          0





          $begingroup$

          I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".



          You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).






          share|cite|improve this answer









          $endgroup$



          I don't understand your question. You say, first, "the inverse image of each set open in Y is open in X" and then ask "what if it is also open in Y?" What does "it" refer to? Lets say that you have a set, U, in X that is mapped into set V in Y: f(U)= V. You have already said "each set open in Y" so that V is necessarily open. U is not in Y (unless you are talking about the special case X= Y) so can't be "open in Y".



          You also talk about X and Y themselves being "open" and "closed" which has nothing to do with "continuity". This definition talks about subsets of X and Y, not X and Y themselves (and any topological space is necessarily both open and closed as a subset of itself).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 15:25









          user247327user247327

          11.1k1515




          11.1k1515












          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11


















          • $begingroup$
            open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
            $endgroup$
            – Peter Long
            Dec 16 '18 at 16:11
















          $begingroup$
          open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
          $endgroup$
          – Peter Long
          Dec 16 '18 at 16:11




          $begingroup$
          open in Y is for me this i.imgur.com/zlR9jMS.png - is this a continuity map ?
          $endgroup$
          – Peter Long
          Dec 16 '18 at 16:11


















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