How do I show that $\text{dist}(x,A)=\varepsilon + \text{dist}(x,B_{\varepsilon}(A))$, if $\text{dist}(x,A)\gt \varepsilon$?

Question

Let $(X,d)$ be a metric space and $A\subset X$ a non-empty subset. Define $B_{\varepsilon}(A):=\{y \in X: \text{dist}(y,A)\leq\varepsilon\}$. I want to prove that if $x\in X$ with $\text{dist}(x,A)\gt \varepsilon$, then

$$\text{dist}(x,A)=\varepsilon + \text{dist}(x,B_{\varepsilon}(A)).$$

I already showed $\text{dist}(x,A)\leq\varepsilon + \text{dist}(x,B_{\varepsilon}(A))$, but I struggle showing $\text{dist}(x,A)\geq\varepsilon + \text{dist}(x,B_{\varepsilon}(A)).$

Answer 1

You have difficulty with this because it is false! Let $X=\mathbb N$ with the discrete metric. Let $A=\{1\}$. Then $B_{\epsilon}(A)=A$ for $0<\epsilon <1$. Take $x=2$ to get a counterexample.