Let $X$ be a metric space with metric $d$.Then $d:X \times X\rightarrow \mathbb R$ is continuous?

Please Check the following proof-

We'll try to show this via sequential crieria of continuity.Let$(x,y)\in X\times X$ be any arbitrary point.We need to show that $d$ is continuous at $(x,y)$.

Let $<(x_n,y_n)>$ be a sequence in $X\times X$ converges to $(x,y)\in X\times X$.Now we'll show that $<d(x_n,y_n)>$ converges to $d(x,y)$.

Since,$\lim_{n\rightarrow {\infty}}(x_n,y_n)=(x,y)$.Therefore,$\lim_{n\rightarrow {\infty}}x_n=x$ &$\lim_{n\rightarrow {\infty}}y_n=y\implies d(x_n,x)<\frac{\epsilon}{2}\forall n>m_1$ and $d(y_n,y)<\frac{\epsilon}{2}\forall n>m_2$.

Now,$d(x_n,y_n)\leq d(x_n,x)+d(x,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n) \implies d(x_n,y_n)-d(x,y)\leq d(x_n,x)+d(y,y_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

i.e$d(x_n,y_n)-d(x,y)<\epsilon$ or $\lim_{n\rightarrow{\infty}}d(x_n,y_n)=d(x,y)$ .

Hence,$\lim_{n\rightarrow{\infty}}(x_n,y_n)=(x,y)\implies \lim_{n\rightarrow{\infty}}d(x_n,y_n)=d(x,y)$. Hence,$d$ is continuous on $X$.

  • $\begingroup$ looks good to me $\endgroup$
    – gt6989b
    Nov 15, 2017 at 16:46
  • $\begingroup$ @gt6989b:My professor told me that it is wrong $\endgroup$
    – P.Styles
    Nov 15, 2017 at 17:05

2 Answers 2


Note that $|d(x,y) - d(w,z)|\leqslant d(x,w) + d(y,z)$, which some people call the parallelogram law. You used this in your proof, in fact.


You need to prove that $d(x,y) - d(x_n,y_n)\leq d(x_n,x)+d(y,y_n)$ holds too, as you need to consider the absolute value of both sides. This can be done in a similar way as you did for the first part.

Proving $d(x,y) - d(x_n,y_n)\leq d(x_n,x)+d(y,y_n)$ and $ d(x_n,y_n) - d(x,y) \leq d(x_n,x)+d(y,y_n)$ is enough to conclude that $|d(x,y) - d(x_n,y_n)|\leq d(x_n,x)+d(y,y_n)$

Also don't forget to write absolute value when necessary. For example you need to prove that $|d(x_n,y_n)-d(x,y)|<\epsilon$ to conclude continuity.


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