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So given a set $Y$ of a topological space $X$ I ask to prove or to disprove if the identity $$ \operatorname{cl}\big(\operatorname{cl} Y\setminus\{x_0\}\big)=\operatorname{cl}Y $$ holds when $x_0$ is an accumulation point for $Y$.

I try to prove the statement as follows. So first of all I observed that $$ \operatorname{cl}Y\setminus\{x_0\}\subseteq\operatorname{cl}Y $$ and thus I conclude that $$ \operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)\subseteq\operatorname{cl}Y $$ Moreover I observed that if $x_0$ is an accumulation point for $Y$ then any neighborhood $V$ of $x_0$ is not disjoint from $Y\setminus\{x_0\}$ and thus from $\operatorname{cl} Y\setminus\{x_0\}$, that is more explicitly I observed that $$ \emptyset\neq V\cap\big(Y\setminus\{x_0\}\big)\subseteq V\cap\big(\operatorname{cl}Y\setminus\{x_0\}\big) $$ for any $V\in\mathcal V(x_0)$ and thus I conclude that $x_0$ is adherent to $\operatorname{cl}Y\setminus\{x_0\}$, that is $$ x_0\in\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big) $$ Now the clausure contains all accumulation points so that an element $x$ of $\operatorname{cl} Y$ can be equal to $x_0$ or can be different to $x_0$: so in the first case the above argumentations show that $x$ is an element of $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$ whereas obviously in the second case $x$ is an element of $\operatorname{cl} Y\setminus\{x_0\}$ and so trivially an element of $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$ and thus I finally concluded that $$ \operatorname{cl}Y\subseteq\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big) $$

So effectively the statement is true when $x_0$ is an accumulation point for $Y$ so that I ask iif the argumentations I gave are correct; however it seems to me that this if also true when $x_0$ is an accumulation point for $\operatorname{cl} Y$ and so I also ask clarifications about; finally I would like to understand if the identity generally holds, that is I would like to know if it holds when $x$ is not an accumulation point for $Y$ or for $\operatorname{cl} Y$. So could someone help me, please?

Antonio Maria Di Mauro
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    If $x_0$ is an accumulation point of $\operatorname{cl}Y$, just apply the result to $\operatorname{cl}Y$ to conclude that $$\operatorname{cl}\big((\operatorname{cl}Y)\setminus{x_0}\big)=\operatorname{cl}\big((\operatorname{cl}\operatorname{cl}Y)\setminus{x_0}\big)=\operatorname{cl}\operatorname{cl}Y=\operatorname{cl}Y,.$$ Now let $X={0}\cup{2^{-n}:n\in\Bbb N}$, and let $Y=X\setminus{0}$; $1=2^0\in Y$, but $1$ is not an accumulation point of $Y$ or of $\operatorname{cl}Y=X$, and ... – Brian M. Scott May 01 '22 at 20:20
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    ... $\operatorname{cl}\big((\operatorname{cl}Y)\setminus{1}\big)=\operatorname{cl}\big(X\setminus{1}\big)=X\setminus{1}\ne\operatorname{cl}Y$. – Brian M. Scott May 01 '22 at 20:20
  • @BrianM.Scott Okay, however the result is true when $x_0$ is an accumulation point for $Y$ or for $\operatorname{cl}Y$, right? – Antonio Maria Di Mauro May 01 '22 at 21:55

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As Brian M. Scott show in the comments above the result is generally false: e.g. given the set $$ X:=\{0\}\cup\{2^{-n}:n\in\Bbb N\} $$and then the subset $$ Y:=X\setminus\{0\} $$ it is not hard to show that $$ \operatorname{cl}\big((\operatorname{cl}Y)\setminus\{1\}\big)=\operatorname{cl}(X\setminus\{1\})=X\setminus\{1\}\neq X=\operatorname{cl}Y $$ with respect the usual topology on $\Bbb R$.

However the result is true for accumulations point of $Y$ and of $\operatorname{cl} Y$ to since if it is true for accumulation points of $Y$ then $$ \operatorname{cl}\big((\operatorname{cl}Y)\setminus\{x_0\}\big)=\operatorname{cl}\big((\operatorname{cl}\operatorname{cl}Y)\setminus\{x_0\}\big)=\operatorname{cl}\operatorname{cl}Y=\operatorname{cl}Y\,. $$ for any accumulation point $x_0$ of $\operatorname{cl}Y$.

Antonio Maria Di Mauro
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