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

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

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 What is your question exactly? – martini Oct 4 '12 at 11:40 prove the above equality – Mathematics Oct 4 '12 at 11:41

## closed as not a real question by Noah Snyder, Norbert, Thomas, tomasz, J. M.Oct 7 '12 at 13:25

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

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 $x\in P'$.

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 aren't $U$ is an open neighborhood of $x$ instead of $y$? – Mathematics Oct 4 '12 at 11:43 @Mathematics: It’s both: it’s an open neighborhood of every point that it contains. – Brian M. Scott Oct 4 '12 at 11:44 o i see. you mean $U$ is an open interval of $y$ in set clP – Mathematics Oct 4 '12 at 11:48 @Mathematics: No, not at all. First, ‘interval’ may not have meaning, since the space may not be a linearly ordered space. Secondly, $U$ is not contained in $\cl P$. – Brian M. Scott Oct 4 '12 at 11:50 I don't quite get why $U$ is the neighborhood of y – Mathematics Oct 4 '12 at 11:52
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 $(\operatorname{cl}P)'=X\ne\emptyset$.