About limit (or accumulation) points and delta neighborhood

Question

I have a problem with limit points when related to closed sets. Supposing that we have a circle (a set $S$) as shown in the image below. A closed set is said to be a set that contains all of its limit(or accumulation) points. However, the set $S$ shown in the picture which lies on the complex plane does not contain point A, point A is supposed to be an accumulation point since its deleted $\delta$ neighborhood shares common points with the set $S$ and supposing that the set $S$ includes both the inside and the border points of the green circle it is supposed to be a closed set(right?). So where is my misunderstanding of the closed sets and accumulation points? (Edit: is the $\delta$ here supposed to be the same as the one from the real analysis? ). Thanks in advance for any help.

Answer 1

The point $A$ is not an accumulation point of $S$. Yes, there is some $\delta>0$ such that the $D_\delta(A)\setminus\{A\}$ intersects $S$. But asserting that $A$ is an accumulation point of $S$ means that this holds for every $\delta>0$. And that's not the case here. Just take some $\delta$ smaller than the distance from $A$ to the center of $S$ minus the radius of $S$.

Answer 2

You are misinterpreting the definition of accumulation point.

Definition

Given a metric space $(X,d_{X})$ and a set $S\subseteq X$, we say that $x\in X$ is an accumulation point of $S$ if, and only if, for every $\varepsilon > 0$ there corresponds a point $p\in S$ such that $0 < d_{X}(x,p) < \varepsilon$.

Comments

In the present example, there corresponds an open ball $B_{\delta}(p)$ which is contained in $S^{c}$, hence $p$ is an exterior point. Therefore it cannot be an accumulation point.