Prove $f: X \to \mathbb{R}_{\geq 0}$ where $f(x) = d(x, x_o)$ is a continuous function

Question

I've read here: Proving that a sphere is a closed set that the function $f: X \to \mathbb{R}_{\geq 0}$ where $f(x) = d(x, x_o)$ is a continuous function and that it can be proven continuous via the triangle inequality. Before I attempt this proof, the answer in the link provided mentioned it's a map from $X$ to $\mathbb{R}$ , but would it be better to say that it's a map from $X$ to $\mathbb{R}_{\geq 0}$ due to distance function's outputs being non-negative?

Let $X$ be a metric space. If $x, x_o \in X$, consider the function $f(x) = d(x, x_o)$ which defines the map $f: X \to \mathbb{R}_{\geq 0}$.

$f$ is continuous at some $x \in X$ if $\forall \epsilon > 0, \exists \delta > 0 : d(x, y) < \delta \Rightarrow d(f(x), f(y)) < \epsilon$ where $y \in X$

suppose we have some $x, y \in X$, some $\epsilon$, and let's choose $\delta = \epsilon$

$d(x, y) < \delta$

$\Rightarrow d(x, y) < \epsilon\:\:\:\:\:\:\:\:(1)$

Consider a third element $x_o \in X$

From the triangle inequality, we have ...

$d(y, x_o) \leq d(x, x_o) + d(x, y)$

$\Rightarrow d(y, x_o) - d(x, x_o) \leq d(x, y)\:\:\:\:\:\:\:\:(2)$

From $(1)$ and $(2)$, we have...

$d(y, x_o) - d(x, x_o) \leq d(x, y) < \epsilon$

$\Rightarrow d(y, x_o) - d(x, x_o) < \epsilon$

we can now substitute in the function

$\Rightarrow f(y) - f(x) < \epsilon$

From the triangle inequality, we also have ...

$d(x, x_o) \leq d(y, x_o) + d(x, y)$

$\Rightarrow d(x, x_o) - d(y, x_o) \leq d(x, y)\:\:\:\:\:\:\:\:(3)$

From $(1)$ and $(3)$, we have...

$d(x, x_o) - d(y, x_o) \leq d(x, y) < \epsilon$

$\Rightarrow d(x, x_o) - d(y, x_o) < \epsilon$

we can now substitute in the function

$\Rightarrow f(x) - f(y) < \epsilon$

$f(y) - f(x)$ and $f(x) - f(y)$ are both less than $\epsilon$

$\Rightarrow |f(x) - f(y)| < \epsilon$

$\Rightarrow d(f(x), f(y)) < \epsilon$

$\Rightarrow$ for any $\epsilon$, we can always find some $\delta$ equal to $\epsilon$ which implies the $\epsilon - \delta$ definition of continuity at any point of $X$

$\Rightarrow f$ is continuous

Answer 1

As @Damian Pavlyshyn suggested, the function $f(x)= d(x,x_0)$ is not only continuous but also satisfies the Lipschitz condition with constant $1.$ Indeed by triangle inequality we have $$d(x,x_0)\le d(x,y)+d(y,x_0)$$ Therefore $$f(x)-f(y)=d(x,x_0)-d(y,x_0)\le d(x,y)$$ Switching $x\leftrightarrow y$ gives $$f(y)-f(x)\le d(y,x)=d(x,y)$$ Hence $$|f(x)-f(y)|\le d(x,y)$$ If one still insists on $\varepsilon$-$\delta$ proof then take $\delta=\varepsilon.$

Answer 2

it is simpler to use sequences. let ${x_n}$ be sequence in a metric space $X$. And $x_n\rightarrow x$. Then, we need to show that $d(x_n,x)\rightarrow 0$. $(\forall \epsilon)(\exists N)(n>N\rightarrow d(x_n,x)<\epsilon)$ and $n, N$ are natural numbers. but this is exactly the definition of a convergent sequence: every ball $B_\epsilon(x)$ contains all but a finite number of points of the sequence. Take $N>1/\epsilon$ and this completes the proof.