For metric space $(X, d)$ and fixed $a\in X$, prove function from induced topology to $\Bbb R$ defined by $f(x) = d(a, x)$ is continuous.

Question

Let the induced topology be $(X, \tau)$. Let $(x, y) \subset \Bbb R$. I'm required to prove that $f^{-1}((x, y))$ is open in $\tau$

$f^{-1}((x, y)) = \{p \in X : x < d(p, a) < y \}$

My thinking is to show this is a union of open balls. Don't know if this is allowed, but for each $p \in f^{-1}((x, y))$, find a $\delta > 0$ such that $B_\delta(p) \subseteq f^{-1}((x, y)) $

Am I going in the right direction with this? Not sure how to proceed regardless.

Answer 1

It might be easier to prove it directly as in:

$f(x) = d(x,a) \le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$ we get $|f(x)-f(y)| \le d(x,y)$ from which continuity follows.

Since $f$ is continuous, $f^{-1} ((x,y))$ is open.

Answer 2

You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^{-1}((x,y))$ is an element of an open ball that is contained in $f^{-1}((x,y))$.

Let $p$ be an arbitrary element of $f^{-1}((x,y))$. Let $r = \min\left\{ \frac{|x - d(p,a)|}{2}, \frac{|y - d(p,a)|}{2} \right\}$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^{-1}((x,y))$ and hence we conclude that $f^{-1}((x,y))$ is open.