Suppose I have a set of points, a closed ball and a point $\vec a$ that is isolated, all in $\mathbb{R}^n$. So I get this set, say $A$, by union of that closed ball and this isolated point $\vec a$ that's not in the closed ball.
Is this a closed set? By the definition of my book, this seems closed because the book says a set is closed if for all $\vec x \not \in$ the set, $\vec x$ is an exterior point.
But I think this isolated point is also a limit point. This is where I get confused too. By definition an isolated point is a point that is not limit point. I don't get this definition. Couldn't my isolated point be a limit point. I can think of a sequence of points in $A$ with limit in $A$, ie. the isolated point $\vec a$.
So, is this isolated point also a limit point and is this set $A$ closed?
Best Answer
A set $C$ in a topological space $X$ is closed if for every $x\in X\setminus C$, there exists an open set $U$ such that $x\in U \subseteq X\setminus C$.
A point $x$ in a subspace $S$ of a topological space $X$ is an isolated point of $S$ if there exists an open set $U$ such that $U\cap S = \{x\}$.
A point $x$ is a limit point of a subspace $S$ of a topological space $X$ if every open set $U$ which contains $x$ satisfies $U\cap (S\setminus\{x\})\ne\emptyset$.
Let's notice something right away:
Proof: Since $x\in S$ by hypothesis, we have that $x$ is an isolated point of $S$ iff there exists $U$ open such that $U\cap S=\{x\}$ iff there exists $U$ open containing $x$ such that $U\cap(S\setminus\{x\})=\emptyset$ iff $x$ is not a limit point of $S$. $\square$
So if we have a closed ball $B$ and a point $x$ in $\mathbb{R}^n\setminus B$, then the union $A:=B\cup\{x\}$ is a closed set, $x$ is an isolated point of $B$, and $x$ is not a limit point of $B$.