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Some denotations:

Let $E$ be a non-empty subset of $\mathbb R^d$,an isolated point x of $E$ is a point in $E$ that $\exists r > 0$, denote $B(x,r)=${$a\in \mathbb R^d:|a-x|<r $}, then $B(x,r)\cap E = ${$x $}.

Denote $E^i$ as the subset of all isolated points of $E$.

Denote $E_0=E,E_n=E_{n-1}\setminus E_{n-1}^i,n = 1,2,3,\dots,F = \cap_{n=1}^{\infty}E_n $. Call points in $E_{n}\setminus E_{n+1}$ n-order points.

My question is: If $F$ is not empty,then must $F^i$ be empty?

Here are some easy facts:

  1. If $x_1$ is a n-order point of $E$,then $\exists \delta >0$, points in $(B(x_1,\delta)\cap E)\setminus $ {$x_1$} are those with
    order less than n.

  2. If there exist n-order points,then there exist n-1-order points.

Assume $F^i$ is not empty,then let $x\in F^i$. Since $x\in F$, $x$ is not isolated in every $E_n$. Also there exists $\delta >0$,points in $(B(x,\delta)\cap E)\setminus $ {$x$} are points with finite order and there are points with arbitrarily large order.

I would be rather appreciated if someone may help me with this question!

DTK
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  • The set $E\setminus E^i$ is called the derived set of $E$ and denoted by $E'$. See https://en.wikipedia.org/wiki/Derived_set_(mathematics). – user14111 Apr 14 '24 at 10:27
  • The short answer to your question is "no". The process of removing isolated points can be continued much farther. It was this that led Georg Cantor to introduce transfinite (ordinal) numbers to mathematics. Let $E_\omega=F$, and follow it with $E_{\omega+1}=F'=F-F^i$, then $E_{\omega+2}=E_{\omega+1}'$, etc. etc. ETC. – user14111 Apr 14 '24 at 10:33
  • @user14111 Thanks for that! And could you please give a counter example? – DTK Apr 14 '24 at 11:29
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    could you please give a counter example? --- For ${\mathbb R}^1,$ see the various sets given in my answer to Intuition for $\omega^\omega$. – Dave L. Renfro Apr 14 '24 at 15:19
  • @DaveL.Renfro Thanks for your comment! I can understand your construction. Actually, just scale the set $\omega ^\omega$ that defined in your answer into the interval $[0,1]$(by an appropriate map) and that's a counter example. – DTK Apr 14 '24 at 16:06

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