How to define limit operations in general topological spaces? Are nets able to do this?is a net stronger than a transfinite sequence for characterizing topology?Permitted value of epsilon in Discrete Metric SpaceWhy are topological spaces interesting to study?Define $f(y)=d(x_0,y)$, prove that $f$ is continuous.Finite point set has limit points for general topological spaces?Is the uniform limit of continuous functions continuous for topological spaces?Can we define the concept of limit without a topology?Equivalent definition of limit of a function (Reference request)Can limits be defined in a more algebraic way, instead of using the completely analytic $delta$-$epsilon$ definition?Is Wikipedia correct about bounded sets?

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How to define limit operations in general topological spaces? Are nets able to do this?


is a net stronger than a transfinite sequence for characterizing topology?Permitted value of epsilon in Discrete Metric SpaceWhy are topological spaces interesting to study?Define $f(y)=d(x_0,y)$, prove that $f$ is continuous.Finite point set has limit points for general topological spaces?Is the uniform limit of continuous functions continuous for topological spaces?Can we define the concept of limit without a topology?Equivalent definition of limit of a function (Reference request)Can limits be defined in a more algebraic way, instead of using the completely analytic $delta$-$epsilon$ definition?Is Wikipedia correct about bounded sets?













4












$begingroup$


I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.



For $f:mathbbRto mathbbR$,
$
lim_xto x_0f(x)=a
$

means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.



Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?










share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.



    For $f:mathbbRto mathbbR$,
    $
    lim_xto x_0f(x)=a
    $

    means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.



    Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?










    share|cite|improve this question











    $endgroup$














      4












      4








      4


      1



      $begingroup$


      I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.



      For $f:mathbbRto mathbbR$,
      $
      lim_xto x_0f(x)=a
      $

      means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.



      Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?










      share|cite|improve this question











      $endgroup$




      I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.



      For $f:mathbbRto mathbbR$,
      $
      lim_xto x_0f(x)=a
      $

      means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.



      Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?







      real-analysis general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 14:17









      YuiTo Cheng

      2,0492637




      2,0492637










      asked Mar 17 at 11:12









      EEEBEEEB

      53138




      53138




















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          The notion of limit is well-defined for any topological space, even non-metric ones.



          Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.



          But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.






          share|cite|improve this answer











          $endgroup$








          • 4




            $begingroup$
            Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
            $endgroup$
            – chi
            2 days ago










          • $begingroup$
            @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
            $endgroup$
            – TheSilverDoe
            2 days ago


















          5












          $begingroup$

          The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.



          Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            But the question was about limits, not continuity.
            $endgroup$
            – Paul Sinclair
            2 days ago


















          4












          $begingroup$

          A net is a function from an directed set $(I, le)$ (say) to a space $X$.



          $f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.



          The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.



          I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.



          If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.






          share|cite|improve this answer











          $endgroup$












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            3 Answers
            3






            active

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            The notion of limit is well-defined for any topological space, even non-metric ones.



            Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.



            But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.






            share|cite|improve this answer











            $endgroup$








            • 4




              $begingroup$
              Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
              $endgroup$
              – chi
              2 days ago










            • $begingroup$
              @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
              $endgroup$
              – TheSilverDoe
              2 days ago















            7












            $begingroup$

            The notion of limit is well-defined for any topological space, even non-metric ones.



            Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.



            But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.






            share|cite|improve this answer











            $endgroup$








            • 4




              $begingroup$
              Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
              $endgroup$
              – chi
              2 days ago










            • $begingroup$
              @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
              $endgroup$
              – TheSilverDoe
              2 days ago













            7












            7








            7





            $begingroup$

            The notion of limit is well-defined for any topological space, even non-metric ones.



            Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.



            But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.






            share|cite|improve this answer











            $endgroup$



            The notion of limit is well-defined for any topological space, even non-metric ones.



            Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.



            But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 17 at 14:46









            psmears

            71149




            71149










            answered Mar 17 at 13:09









            TheSilverDoeTheSilverDoe

            4,272114




            4,272114







            • 4




              $begingroup$
              Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
              $endgroup$
              – chi
              2 days ago










            • $begingroup$
              @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
              $endgroup$
              – TheSilverDoe
              2 days ago












            • 4




              $begingroup$
              Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
              $endgroup$
              – chi
              2 days ago










            • $begingroup$
              @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
              $endgroup$
              – TheSilverDoe
              2 days ago







            4




            4




            $begingroup$
            Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
            $endgroup$
            – chi
            2 days ago




            $begingroup$
            Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
            $endgroup$
            – chi
            2 days ago












            $begingroup$
            @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
            $endgroup$
            – TheSilverDoe
            2 days ago




            $begingroup$
            @chi Yes, thank you for this comment ! Indeed the precision can be usefull.
            $endgroup$
            – TheSilverDoe
            2 days ago











            5












            $begingroup$

            The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.



            Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              But the question was about limits, not continuity.
              $endgroup$
              – Paul Sinclair
              2 days ago















            5












            $begingroup$

            The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.



            Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              But the question was about limits, not continuity.
              $endgroup$
              – Paul Sinclair
              2 days ago













            5












            5








            5





            $begingroup$

            The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.



            Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.






            share|cite|improve this answer









            $endgroup$



            The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.



            Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 17 at 13:11









            Kavi Rama MurthyKavi Rama Murthy

            68.6k53169




            68.6k53169







            • 1




              $begingroup$
              But the question was about limits, not continuity.
              $endgroup$
              – Paul Sinclair
              2 days ago












            • 1




              $begingroup$
              But the question was about limits, not continuity.
              $endgroup$
              – Paul Sinclair
              2 days ago







            1




            1




            $begingroup$
            But the question was about limits, not continuity.
            $endgroup$
            – Paul Sinclair
            2 days ago




            $begingroup$
            But the question was about limits, not continuity.
            $endgroup$
            – Paul Sinclair
            2 days ago











            4












            $begingroup$

            A net is a function from an directed set $(I, le)$ (say) to a space $X$.



            $f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.



            The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.



            I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.



            If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.






            share|cite|improve this answer











            $endgroup$

















              4












              $begingroup$

              A net is a function from an directed set $(I, le)$ (say) to a space $X$.



              $f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.



              The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.



              I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.



              If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.






              share|cite|improve this answer











              $endgroup$















                4












                4








                4





                $begingroup$

                A net is a function from an directed set $(I, le)$ (say) to a space $X$.



                $f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.



                The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.



                I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.



                If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.






                share|cite|improve this answer











                $endgroup$



                A net is a function from an directed set $(I, le)$ (say) to a space $X$.



                $f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.



                The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.



                I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.



                If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered Mar 17 at 13:40









                Henno BrandsmaHenno Brandsma

                113k348123




                113k348123



























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