Checking Continuity of functions

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Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions







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    If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
    – Lord Shark the Unknown
    Aug 19 at 9:10














up vote
2
down vote

favorite












Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions







share|cite|improve this question


















  • 3




    If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
    – Lord Shark the Unknown
    Aug 19 at 9:10












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions







share|cite|improve this question














Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 19 at 9:10

























asked Aug 19 at 9:09







user501655














  • 3




    If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
    – Lord Shark the Unknown
    Aug 19 at 9:10












  • 3




    If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
    – Lord Shark the Unknown
    Aug 19 at 9:10







3




3




If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
– Lord Shark the Unknown
Aug 19 at 9:10




If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
– Lord Shark the Unknown
Aug 19 at 9:10










3 Answers
3






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1
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If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if



(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$



or



(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).



In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.






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    up vote
    0
    down vote













    What is your definition for $l=limlimits_xto 1f(x)$ if it exists?



    If this is it:



    $$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$



    then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Continuity is a topological concept.



      If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.



      So that is the criterion that should be handled here.



      Observe that limits do not play a part in it.



      In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).



      Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.




      It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.



      To get more view on this you could take a look at this answer.






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

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        up vote
        1
        down vote













        If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if



        (1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$



        or



        (2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).



        In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.






        share|cite|improve this answer
























          up vote
          1
          down vote













          If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if



          (1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$



          or



          (2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).



          In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.






          share|cite|improve this answer






















            up vote
            1
            down vote










            up vote
            1
            down vote









            If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if



            (1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$



            or



            (2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).



            In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.






            share|cite|improve this answer












            If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if



            (1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$



            or



            (2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).



            In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 19 at 9:55









            Hans Lundmark

            33.2k564109




            33.2k564109




















                up vote
                0
                down vote













                What is your definition for $l=limlimits_xto 1f(x)$ if it exists?



                If this is it:



                $$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$



                then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.






                share|cite|improve this answer
























                  up vote
                  0
                  down vote













                  What is your definition for $l=limlimits_xto 1f(x)$ if it exists?



                  If this is it:



                  $$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$



                  then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.






                  share|cite|improve this answer






















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    What is your definition for $l=limlimits_xto 1f(x)$ if it exists?



                    If this is it:



                    $$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$



                    then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.






                    share|cite|improve this answer












                    What is your definition for $l=limlimits_xto 1f(x)$ if it exists?



                    If this is it:



                    $$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$



                    then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 19 at 9:29









                    Scientifica

                    4,90121231




                    4,90121231




















                        up vote
                        0
                        down vote













                        Continuity is a topological concept.



                        If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.



                        So that is the criterion that should be handled here.



                        Observe that limits do not play a part in it.



                        In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).



                        Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.




                        It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.



                        To get more view on this you could take a look at this answer.






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          Continuity is a topological concept.



                          If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.



                          So that is the criterion that should be handled here.



                          Observe that limits do not play a part in it.



                          In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).



                          Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.




                          It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.



                          To get more view on this you could take a look at this answer.






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Continuity is a topological concept.



                            If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.



                            So that is the criterion that should be handled here.



                            Observe that limits do not play a part in it.



                            In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).



                            Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.




                            It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.



                            To get more view on this you could take a look at this answer.






                            share|cite|improve this answer












                            Continuity is a topological concept.



                            If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.



                            So that is the criterion that should be handled here.



                            Observe that limits do not play a part in it.



                            In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).



                            Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.




                            It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.



                            To get more view on this you could take a look at this answer.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 19 at 9:35









                            drhab

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