A question about proving the derived set of closure P =derived set of P [closed]

Question

As the topic, proving that $(\operatorname{cl} P)'=P'$, where $P'$ means the derived set.

Answer 1

The result is not true in completely general spaces. However, it is true in $T_1$ spaces, so I’ll assume that we’re working in a $T_1$ space.

$\newcommand{\cl}{\operatorname{cl}}$It should be clear that $P'\subseteq(\cl P)'$. To show the opposite inclusion, suppose that $x\in(\cl P)'$. Then every open neighborhood of $x$ contains a point of $(\cl P)\setminus\{x\}$, and we want to show that every open neighborhood of $x$ contains a point of $P\setminus\{x\}$.

Suppose not; then $x$ has an open neighborhood $U$ such that $U\cap P\subseteq\{x\}$. Then either $U\cap P=\varnothing$, or $U\cap P=\{x\}$. If $U\cap P=\varnothing$, then $U\cap\cl P=\varnothing$ (why?), and therefore $x\notin(\cl P)'$; this is a contradiction, so we must have $U\cap P=\{x\}$. In other words, $x$ is an isolated point of $P$. By hypothesis, however, $U$ contains some point $y\in\cl P\setminus\{x\}$. $U\cap P=\{x\}$, so $y\notin P$, and therefore $y\in P'\setminus P$. Now let $V=U\setminus\{x\}$; $\{x\}$ is closed, since the space is $T_1$, so $V$ is an open neighborhood of $y$. But $V\cap P=\varnothing$, contradicting the fact that $y\in\cl P$, so this case is also impossible. Thus, $x$ cannot have an open neighborhood $U$ such that $u\cap P\subseteq\{x\}$, so every open neighborhood of $x$ contains a point of $P\setminus\{x\}$, and $x\in P'$.

Answer 2

If $x$ is an accumulation point of $P$, then it is also one in $\operatorname{cl}P$, hence $P'\subseteq (\operatorname{cl}P)'$ is clear.

The other direction is not true in general. Consider the two-point pace $X=\{x,y\}$ with indiscrete topology amd $P=\{x\}$. Then $P'=\{y\}$ and $\operatorname{cl}P=X$, hence $(\operatorname{cl}P)'=X\ne\emptyset$.