help me please proof about contiunity of metric

Question

metric is a continuous function.I need to prove that.But I cannot show that if $|x-a|<\delta$ then $|d(x,y)-d(y,a)| < \varepsilon $. for any metric $d$

$|x-a|<\delta$ is it true? or Should I that $d(x,a)<\delta$. I am confusing

Answer 1

Let $E$ be your metric space. $d:E\times E \to \mathbb R_+$ its distance.

Saying that $d$ is continuous at $X\in E\times E$ is saying that $\forall \varepsilon>0, \exists \alpha >0, \forall Y \in E\times E, D(X,Y)\le \alpha \implies |d(X)-d(Y)|\le \varepsilon$

The problem here is that you didn't give $D$, the metric on $E\times E$. You can define it using $d$ but there are several ways to do so and some might not make $d$ continuous. Your statement is true for $D((x,y),(a,b))=\max\{d(x,a),d(y,b)\}$.

$$\begin{array}{ll} |d(x,y)-d(a,b)| &= |d(x,y)-d(x,b)+d(x,b)-d(a,b)|\\ &\le |d(x,y)-d(x,b)|+|d(x,b)-d(a,b)|\\ &\le d(y,b)+d(x,a)\\ &\le 2\max\{d(x,a),d(y,b)\}\\ |d(x,y)-d(a,b)| &\le 2 D((x,y),(a,b)) \end{array}$$

So you can take $\alpha = \frac{\varepsilon}{2}$.