Is the delta neighborhood of an epsilon neighborhood of a complete metric space the same as the delta + epsilon neighborhood

Question

Let X be a complete metric space. Does there exist a compact subset A that is non empty s.t. the delta neighborhood of the epsilon neighborhood of A is not the same as the epsilon + delta neighborhood?

I start by taking the closed unit interval. I'm thinking of looking at two points that do not coincide. Then take epsilon so the neighborhoods around the points don't touch, but I don't think that works.

(I've been advised to use a finite subset of a Euclidean space as X.)

Answer 1

As long as $|X| > 1$ yes. Let $x \in X$. Since $X$ is finite $$d := \inf_{y \in X\setminus \{x\}} d(x,y) > 0$$ Then $$N_{\frac{3d}{4}}(N_{\frac{3d}{4}}(x)) = N_{\frac{3d}{4}}(x)= \{x\}$$ On the other hand, there is $y \in X$ with $y \neq x$ so that $d(y, x) = d$ so $y \in N_{\frac{3d}{4} + \frac{3d}{4}}(x)$.

Answer 2

I can give you an example in a noneuclidean space.

Take X to be any discrete metric space containing more than one points . It is complete.

Consider any $\{x\}$ it is compact. And if you take $\epsilon =3/4, \delta =1/2$ you get the example.

Now a non-singleton finite subset of an euclidean space is discrete. Only you have to choose $\epsilon$ and $\delta$ carefully.