Metric is continuous, on the right track?

Solution 1:

Let $W$ be an open interval of center $d(x,y)$ and radius $2r$ in $\mathbb R$, that's $$W=(d(x,y)-2r,d(x,y)+2r)$$ Let $U=D(x,r)\subseteq X$ (disc of center $x$ and radius $r$) and $V=D(y,r)\subseteq X$ then for $(u,v)\in U\times V$ we have:

$d(u,v)\leq d(u,x)+d(x,y)+d(y,v)<d(x,y)+2r$ and $d(x,y)\leq d(u,x)+d(u,v)+d(y,v)<d(u,v)+2r$

from which $d(x,y)-2r<d(u,v)<d(x,y)+2r$.

Consequenlty, $(x,y)\in U\times V\subseteq d^{-1}(W)$.

Solution 2:

Suppose $(X,d)$ is a metric space. Then the metric $d_*:(X\times X)\times(X\times X)\rightarrow \mathbb{R}$ defined by \begin{equation} d_*\big((x,y),(x_o,y_o)\big)=\max\{d(x,x_o),d(y,y_o)\} \end{equation}
is a metric on $X\times X$ induce by the metric $d$ on $X$.

Now, let $(x_0,y_0)\in X\times X$ and $\epsilon>0$. If $(x,y)\in X\times X$ and $d_*\big((x,y),(x_0,y_0)\big)<\frac{\epsilon}{2}$. Then \begin{align*} |d(x,y)-d(x_0,y_0)| &=|d(x,y)-d(x_0,y)+d(x_0,y)-d(x_0,y_0)|\\ &\leq |d(x,y)-d(x_0,y)|+|d(x_0,y)-d(x_0,y_0)|\quad (\text{By Triangular Inequality})\\ &\leq d(x,x_0)+d(y,y_0)\quad \quad (\text{By Reverse Triangule Inequality})\\ &\leq \max\{d(x,x_o),d(y,y_o)\}+\max\{d(x,x_o),d(y,y_o)\}\\ &= d_*\big((x,y),(x_0,y_0)\big)+d_*\big((x,y),(x_0,y_0)\big)\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \end{align*} Hence, $d$ is continuous.