How is the discrete metric continuous?

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It is proven that any metric $d$ is continuous.
Consider the metric space $(mathbbR, d)$ where:



$$d:mathbbRtimesmathbbRrightarrow mathbbR$$



$$d(x, y) = begincases
0 & x=y \
1 & xneq y
endcases
$$



Let $x_n rightarrow x=0$ and $y_n rightarrow y=0$. If you take $d(x_n, y_n)rightarrow 1 neq 0=d(x, y)$. This shows that this metric is discontinuous. What is wrong with my reasoning?










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  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    yesterday










  • You write $mathbbR$ but it has two different topologies for this statement. See my answer.
    – Henno Brandsma
    yesterday














up vote
8
down vote

favorite












It is proven that any metric $d$ is continuous.
Consider the metric space $(mathbbR, d)$ where:



$$d:mathbbRtimesmathbbRrightarrow mathbbR$$



$$d(x, y) = begincases
0 & x=y \
1 & xneq y
endcases
$$



Let $x_n rightarrow x=0$ and $y_n rightarrow y=0$. If you take $d(x_n, y_n)rightarrow 1 neq 0=d(x, y)$. This shows that this metric is discontinuous. What is wrong with my reasoning?










share|cite|improve this question























  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    yesterday










  • You write $mathbbR$ but it has two different topologies for this statement. See my answer.
    – Henno Brandsma
    yesterday












up vote
8
down vote

favorite









up vote
8
down vote

favorite











It is proven that any metric $d$ is continuous.
Consider the metric space $(mathbbR, d)$ where:



$$d:mathbbRtimesmathbbRrightarrow mathbbR$$



$$d(x, y) = begincases
0 & x=y \
1 & xneq y
endcases
$$



Let $x_n rightarrow x=0$ and $y_n rightarrow y=0$. If you take $d(x_n, y_n)rightarrow 1 neq 0=d(x, y)$. This shows that this metric is discontinuous. What is wrong with my reasoning?










share|cite|improve this question















It is proven that any metric $d$ is continuous.
Consider the metric space $(mathbbR, d)$ where:



$$d:mathbbRtimesmathbbRrightarrow mathbbR$$



$$d(x, y) = begincases
0 & x=y \
1 & xneq y
endcases
$$



Let $x_n rightarrow x=0$ and $y_n rightarrow y=0$. If you take $d(x_n, y_n)rightarrow 1 neq 0=d(x, y)$. This shows that this metric is discontinuous. What is wrong with my reasoning?







real-analysis continuity metric-spaces






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share|cite|improve this question













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edited yesterday









ervx

9,71831337




9,71831337










asked yesterday









Dylan Zammit

8421316




8421316











  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    yesterday










  • You write $mathbbR$ but it has two different topologies for this statement. See my answer.
    – Henno Brandsma
    yesterday
















  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    yesterday










  • You write $mathbbR$ but it has two different topologies for this statement. See my answer.
    – Henno Brandsma
    yesterday















$x_n=0=y_n$ for all $n$ large.
– Jo'
yesterday




$x_n=0=y_n$ for all $n$ large.
– Jo'
yesterday












You write $mathbbR$ but it has two different topologies for this statement. See my answer.
– Henno Brandsma
yesterday




You write $mathbbR$ but it has two different topologies for this statement. See my answer.
– Henno Brandsma
yesterday










3 Answers
3






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oldest

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up vote
8
down vote



accepted










The fact that $d(x_n,x)to 0$ implies that eventually $d(x_n,x)<1/2$. Thus, eventually the $x_n$'s must all be $0$. The same goes for the $y_n$'s. Therefore, $d(x_n,y_n)$ is eventually $0$ for such sequences.






share|cite|improve this answer



























    up vote
    5
    down vote













    You misrepresent the statement.



    Correct is: let $(X,d)$ be a metric space. This induces a topology $mathcalT_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,mathcalT_d) times (X,mathcalT_d) to mathbbR$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.



    If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).






    share|cite|improve this answer



























      up vote
      2
      down vote













      A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?






      share|cite|improve this answer




















      • Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
        – Dylan Zammit
        yesterday










      • Ok, understood what you mean with the help of ervx's answer :) thanks
        – Dylan Zammit
        yesterday










      Your Answer




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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      8
      down vote



      accepted










      The fact that $d(x_n,x)to 0$ implies that eventually $d(x_n,x)<1/2$. Thus, eventually the $x_n$'s must all be $0$. The same goes for the $y_n$'s. Therefore, $d(x_n,y_n)$ is eventually $0$ for such sequences.






      share|cite|improve this answer
























        up vote
        8
        down vote



        accepted










        The fact that $d(x_n,x)to 0$ implies that eventually $d(x_n,x)<1/2$. Thus, eventually the $x_n$'s must all be $0$. The same goes for the $y_n$'s. Therefore, $d(x_n,y_n)$ is eventually $0$ for such sequences.






        share|cite|improve this answer






















          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          The fact that $d(x_n,x)to 0$ implies that eventually $d(x_n,x)<1/2$. Thus, eventually the $x_n$'s must all be $0$. The same goes for the $y_n$'s. Therefore, $d(x_n,y_n)$ is eventually $0$ for such sequences.






          share|cite|improve this answer












          The fact that $d(x_n,x)to 0$ implies that eventually $d(x_n,x)<1/2$. Thus, eventually the $x_n$'s must all be $0$. The same goes for the $y_n$'s. Therefore, $d(x_n,y_n)$ is eventually $0$ for such sequences.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          ervx

          9,71831337




          9,71831337




















              up vote
              5
              down vote













              You misrepresent the statement.



              Correct is: let $(X,d)$ be a metric space. This induces a topology $mathcalT_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,mathcalT_d) times (X,mathcalT_d) to mathbbR$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.



              If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).






              share|cite|improve this answer
























                up vote
                5
                down vote













                You misrepresent the statement.



                Correct is: let $(X,d)$ be a metric space. This induces a topology $mathcalT_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,mathcalT_d) times (X,mathcalT_d) to mathbbR$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.



                If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).






                share|cite|improve this answer






















                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









                  You misrepresent the statement.



                  Correct is: let $(X,d)$ be a metric space. This induces a topology $mathcalT_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,mathcalT_d) times (X,mathcalT_d) to mathbbR$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.



                  If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).






                  share|cite|improve this answer












                  You misrepresent the statement.



                  Correct is: let $(X,d)$ be a metric space. This induces a topology $mathcalT_d$ on $X$ induced by the metric (the minimal topology where all balls $B(x,r)$ are open). Then the map $d: (X,mathcalT_d) times (X,mathcalT_d) to mathbbR$ is continuous where the left hand side has the product topology of the metric topology with itself and the reals have the usual Euclidean topology.



                  If we use the discrete metric, which induces the discrete topology, then this product topology is also discrete and $d$ is indeed continuous (as is any map on a discrete space, so it's not very informative).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Henno Brandsma

                  96.2k343104




                  96.2k343104




















                      up vote
                      2
                      down vote













                      A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?






                      share|cite|improve this answer




















                      • Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                        – Dylan Zammit
                        yesterday










                      • Ok, understood what you mean with the help of ervx's answer :) thanks
                        – Dylan Zammit
                        yesterday














                      up vote
                      2
                      down vote













                      A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?






                      share|cite|improve this answer




















                      • Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                        – Dylan Zammit
                        yesterday










                      • Ok, understood what you mean with the help of ervx's answer :) thanks
                        – Dylan Zammit
                        yesterday












                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?






                      share|cite|improve this answer












                      A sequence $x_n$ has limit $0$ in the discrete metric if and only if eventually $x_n = 0$. Does this help?







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      Ethan Bolker

                      37.2k54299




                      37.2k54299











                      • Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                        – Dylan Zammit
                        yesterday










                      • Ok, understood what you mean with the help of ervx's answer :) thanks
                        – Dylan Zammit
                        yesterday
















                      • Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                        – Dylan Zammit
                        yesterday










                      • Ok, understood what you mean with the help of ervx's answer :) thanks
                        – Dylan Zammit
                        yesterday















                      Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                      – Dylan Zammit
                      yesterday




                      Why is that so? Isn't $x_n$ a normal sequence such as $frac1n$? In which case $x_nneq 0 forall n$?
                      – Dylan Zammit
                      yesterday












                      Ok, understood what you mean with the help of ervx's answer :) thanks
                      – Dylan Zammit
                      yesterday




                      Ok, understood what you mean with the help of ervx's answer :) thanks
                      – Dylan Zammit
                      yesterday

















                       

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