Continuous function in metric spaces

Question

$P = \mathbb R, \rho(x,y):|x|+|y|$ if $x \ne y $ or $0$ if $x=y$. $f(x): (P,\rho) → (\mathbb R,\rho_1): 0$ if $x∈[-1,1]$ or $1$ if $x∈\mathbb R/[-1,1]$. $\rho_1 (x,y) = \sum_{k=1}^\infty |x_k-y_k|$. Is f(x) continuous function? So far I was looking for sequence, for which $x_n \to x_0$ for $n \to \infty: f(x_n) \to q, q\ne f(x_0)$, but I wasn't sucessful. I guess it is continuous, but I don't know how to prove it rigoriously. Any hints?

Answer 1

Let us show continuity at $x_0 \in P$. Let $\varepsilon > 0$.

First suppose $x_0=0$, then for any $y \in P$ satisfying $\rho(x_0,y)< 1/2$ holds $1/2 > |x_0|+|y|= |y|$, i.e. $y \in ]-1/2,1/2[$ and so $f(y)=f(x_0)$. It follows $\rho_1(f(x_0),f(y)) =\rho_1(f(x_0),f(x_0))= 0< \varepsilon$ which shows that $f$ is continuous at $x_0=0$.

Now assume $|x_0| > 0$. Let $y \in P$ such that $\rho(x_0,y)< |x_0|/2$. We claim that $y = x_0$. Suppose this is not true, then $0<|x_0|\leq |x_0|+|y|=\rho(x_0,y)<|x_0|/2$ which is a contradiction. It follows that for every $y \in P$ such that $\rho(x_0,y)< |x_0|/2$, $\rho_1(f(x_0),f(y))=\rho_1(f(x_0),f(x_0))=0<\varepsilon$ and thus $f$ is continuous at $x_0 \neq 0$.