Example of a continuous function that don't have a continuous extension












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Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










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    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    19 hours ago
















5












$begingroup$



Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    19 hours ago














5












5








5


0



$begingroup$



Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










share|cite|improve this question









$endgroup$





Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.







general-topology continuity






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asked 19 hours ago









AnalyticHarmonyAnalyticHarmony

699313




699313








  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    19 hours ago














  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    19 hours ago








1




1




$begingroup$
The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
$endgroup$
– Melody
19 hours ago




$begingroup$
The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
$endgroup$
– Melody
19 hours ago










3 Answers
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Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



    Now, drawing a picture will make the following obvious:



    Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



    If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Another reason: continuous in the whole line implies locally bounded near every point.



      And another counterexample based in a different idea: $Bbb Q$ is dense in $Bbb R$ with the usual topology. The function
      $$f:Bbb QlongrightarrowBbb R$$
      $$
      f(x) =
      begin{cases}
      0:& x < sqrt2,\
      1:& x > sqrt2,
      end{cases}
      $$

      is continuous (check it) and can't be extended continuously to $Bbb R$.






      share|cite|improve this answer











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        3 Answers
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        3 Answers
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        $begingroup$

        Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



        One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





        BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



          One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





          BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



            One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





            BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






            share|cite|improve this answer











            $endgroup$



            Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



            One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





            BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 18 hours ago

























            answered 18 hours ago









            NazimJNazimJ

            79019




            79019























                2












                $begingroup$

                Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                Now, drawing a picture will make the following obvious:



                Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                  Now, drawing a picture will make the following obvious:



                  Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                  If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                    Now, drawing a picture will make the following obvious:



                    Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                    If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






                    share|cite|improve this answer









                    $endgroup$



                    Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                    Now, drawing a picture will make the following obvious:



                    Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                    If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 17 hours ago









                    MatematletaMatematleta

                    12.1k21020




                    12.1k21020























                        1












                        $begingroup$

                        Another reason: continuous in the whole line implies locally bounded near every point.



                        And another counterexample based in a different idea: $Bbb Q$ is dense in $Bbb R$ with the usual topology. The function
                        $$f:Bbb QlongrightarrowBbb R$$
                        $$
                        f(x) =
                        begin{cases}
                        0:& x < sqrt2,\
                        1:& x > sqrt2,
                        end{cases}
                        $$

                        is continuous (check it) and can't be extended continuously to $Bbb R$.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Another reason: continuous in the whole line implies locally bounded near every point.



                          And another counterexample based in a different idea: $Bbb Q$ is dense in $Bbb R$ with the usual topology. The function
                          $$f:Bbb QlongrightarrowBbb R$$
                          $$
                          f(x) =
                          begin{cases}
                          0:& x < sqrt2,\
                          1:& x > sqrt2,
                          end{cases}
                          $$

                          is continuous (check it) and can't be extended continuously to $Bbb R$.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Another reason: continuous in the whole line implies locally bounded near every point.



                            And another counterexample based in a different idea: $Bbb Q$ is dense in $Bbb R$ with the usual topology. The function
                            $$f:Bbb QlongrightarrowBbb R$$
                            $$
                            f(x) =
                            begin{cases}
                            0:& x < sqrt2,\
                            1:& x > sqrt2,
                            end{cases}
                            $$

                            is continuous (check it) and can't be extended continuously to $Bbb R$.






                            share|cite|improve this answer











                            $endgroup$



                            Another reason: continuous in the whole line implies locally bounded near every point.



                            And another counterexample based in a different idea: $Bbb Q$ is dense in $Bbb R$ with the usual topology. The function
                            $$f:Bbb QlongrightarrowBbb R$$
                            $$
                            f(x) =
                            begin{cases}
                            0:& x < sqrt2,\
                            1:& x > sqrt2,
                            end{cases}
                            $$

                            is continuous (check it) and can't be extended continuously to $Bbb R$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 12 hours ago

























                            answered 16 hours ago









                            Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                            35.4k42972




                            35.4k42972






























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