I have the following problem that I've been working on (and I think I have done).
Let $(X,d)$ be a metric space and fix $x_{0}\in X$. Show that the function $f:X\to\mathbb{R}$ defined by $f(x)=d(x,x_{0})$ is continuous.
My attempt: To show that a map $g:X\to Y$ between metric spaces is continuous, we must show that $\{x_{n}\}\to x_{0}$ in $X$ implies that $\{f(x_{n})\}\to f(x_{0})$ in $Y$. For our map $f:X\to \mathbb{R}$, assume that $\{x_{n}\}$ is a sequence in $X$ that converges to $x_{0}$. Then $f(x_{n})=d(x_{n},x_{0})$ so that, in particular, $f(x_{0})=d(x_{0},x_{0})=0$. Thus, it will suffice to show that $\{d(x_{n},x_{0})\}\to 0$ in $\mathbb{R}$. This sequence converges to zero if, for each $\epsilon>0$, there exists an $N$ such that $d(x_{n},x_{0})\in B(0,\epsilon)$ or, equivalently, such that $|d(x_{n},x_{0})|<\epsilon$. But since $\{x_{n}\}\to x_{0}$, we may always find an $N$ such that $d(x_{n},x_{0})<\epsilon$. Hence, $f$ is continuous.
Does this look okay? It's been a while since I've proven continuity using convergent sequences. Thanks for any help!
Yes, it is correct. But I think that it would be more natural to use the fact that$$\forall x,y,z\in X :\lvert d(x,y)-d(x,z)\rvert\le d(y,z).$$ So, in order to prove that $f$ is continuous by the definition of continuity, all it takes is to take $\delta=\varepsilon$.