Let $\tau$ be a indiscrete topology on $X$, such that $|X|\geqslant 2$.
$A'=\begin{cases}
X, & \mbox{ if }|A|\geqslant 2\\
\emptyset, & \mbox{ if }|A|=\emptyset\\
X\setminus A & \mbox{ if }|A|=1 \end{cases},
$
A'– stands for the set of accumulation points A
I am having a hard time understanding this example.
Question:
How can $X\setminus A$ be the set of accumulation points for $A$ if it does not intersect $A$? Should it not be $X$?
Best Answer
Given a topological space ($X$,$\tau$) and a subset $S$ of $X$, an point $x\in X$ is an accumulation point of $S$ if $$\forall U\in\tau(x),\exists s\in S, s\neq x\land s\in U,$$ i.e, every open neighborhood of $x$ (I denote $\tau(x)$ by the set of open neighborhoods of $x$) contains a point of $S$ different than $x$.
Now, the indiscrete topology on $X$ is $\tau=\{\emptyset,X\}$. Let $A\subset X$ s.t $|A|=1$, i.e, $A=\{a\}$ for some $a\in X$.
According to the definition, $a$ is not an accumulation point of $A$, because there's no point in $A$ besides $a$, so you can't find points of $A$ different than $a$ in any open neighborhood of $a$. This shows that $A'\subset X\backslash\{a\}=X\backslash A$. ($A'$ is called derived set). Then since we're working in the indiscrete topology, you have $X\backslash A\subset A'$ as well, hence $A'=X\backslash A$.