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?






continuity metric-spaces







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  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    40 mins ago














up vote
4
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?






continuity metric-spaces







share|cite|improve this question





















  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    40 mins ago












up vote
4
down vote

favorite









up vote
4
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?






continuity metric-spaces







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?






continuity metric-spaces




continuity metric-spaces






share|cite|improve this question













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asked 45 mins ago









Dylan Zammit

8221316




8221316











  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    40 mins ago
















  • $x_n=0=y_n$ for all $n$ large.
    – Jo'
    40 mins ago















$x_n=0=y_n$ for all $n$ large.
– Jo'
40 mins ago




$x_n=0=y_n$ for all $n$ large.
– Jo'
40 mins ago










2 Answers
2






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3
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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.






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    2
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    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
      38 mins ago










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










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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









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    active

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    up vote
    3
    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
      3
      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
        3
        down vote



        accepted







        up vote
        3
        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 42 mins ago









        ervx

        9,69331337




        9,69331337




















            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
              38 mins ago










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














            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
              38 mins ago










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












            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 41 mins ago









            Ethan Bolker

            37.1k54299




            37.1k54299











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










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
















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










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















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




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












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




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

















             

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