Let (X, d) be metric and $U \subset X$. Prove that for any x ∈ X and any limit point $x_0$ of U, $\lim_{x→x0} d(a,x) =d(a,x_0)$

Question

Problem: Let (X, d) be metric and $U \subset X$. Prove that for any x ∈ X and any limit point $x_0$ of U, $\lim_{x→x0} d(a,x) = d(a,x_0)$.

I know this implies that $\forall \varepsilon >0, \exists \delta >0$ such that $\forall x$ where $\left | x - x_0 \right | < \delta$, implies that $\left | d(a,x) - d(a,x_0) \right | < \epsilon$. However, I don't know how I can prove this just using things like the Triangle Inequality and properties of a metric. Any help would be appreciated! I'm also unsure how limit points of a subset (U) fit into this.

Answer 1

Note that$$(\forall x\in X):\bigl\lvert d(x,a)-d(x_0,a)\bigr\rvert\leqslant d(x,x_0).$$Therefore, given $\varepsilon>0$, if you take $\delta=\varepsilon$, then$$d(x,x_0)<\delta\iff d(x,x_0)<\varepsilon\implies\lvert d(x,a)-d(x_0,a)\bigr\rvert<\varepsilon.$$