Metric space and continuity

We define a map $f:(S,d)→(S',d')$ between 2 metric spaces to be continuous at $x \in S$ if for every sequence ${x_n}$ in $S$ that converges to $x$, the sequence $\{f(x_n)\}$ in $S'$ converges to $f(x)$.We are given that $(S,d)$ is a metric space and $x_0$ is a fixed point in $S$. Define $f:(S,d)→(\Bbb R,d_{std})$ to be $f(x):=d(x,x_0)$.

If we want to prove that $d$ is continuous, we know that $f$ is continuous and $\{f(x_n)\}$ is convergent. So according to the theorem, $d$ has to be convergent. Then we can prove that $d$ is continuous?

Solution 1:

You want to prove that if $x_n \to x$ in $S$, then $d(x_n,x_0) \to d(x,x_0)$ in $\Bbb R$. One does this as follows: let $\epsilon > 0$. By convergence of $(x_n)_{n\geq 1}$ there is $n_0$ large enough such that $d(x_n,x)< \epsilon$ if $n \geq n_0$. Then: $$ |d(x_n,x_0)-d(x,x_0)| \color{red}{\leq} d(x_n,x) < \epsilon $$for all $n \geq n_0$, and so $f$ is continuous (the same $n_0$ works for both sequences). The challenge in this is proving the inequality in red. This can be done using the triangle inequality twice and the definition of absolute value - this task I leave to you.

Solution 2:

Let $(a,b)\in S\text{x} S$. By the axiomes of the distance $d$ you have for a point $(x,y)\in S\text{x} S$

$d(x,y)\le d(x,a)+d(a,b)+d(b,y)\Rightarrow d(x,y)-d(a,b)\le d(x,a)+d(b,y)$

Permuting $(y,b)$ and $(x,a)$ you get $$|d(x,y)-d(a,b)|\le d(x,a)+d(b,y)$$

Hence for a given $\epsilon >0$ you can choose $d(x,a)\le \frac {\epsilon}{2}$ and $d(b,y)\le \frac {\epsilon}{2}$ to verify that $$(x,y)\to (a,b)\Rightarrow |d(x,y)-d(a,b)|\to 0$$

This proves the continuity of $d$ ($d$ has uniform continuity indeed).