For an uncountable subset $E\subset\mathbb{R}^n$, I know the following results are true,

  1. $E'$ is uncountable, closed and perfect.
  2. Atmost countably many points of $E$ are not in $E'$

Question: Does this results hold true for countable and finite subsets.

For finite sets it is easy to see because if $E$ is finite, $E'=\emptyset$. $E'$ is closed and perfect(vacuously true).

For countable sets??

Edits: $E'$ denote set of all limit points of the set $E$.

  • $\begingroup$ What does $E'$ mean here? $\endgroup$ – D. Brogan Jan 8 '19 at 17:00
  • $\begingroup$ $E'$ denote set of all limit points of the set $E$.@D.Brogan $\endgroup$ – StammeringMathematician Jan 8 '19 at 17:05
  • $\begingroup$ Consider $\mathbb Z$ and $\mathbb Q$ in the space $\mathbb R$ of real numbers. $\endgroup$ – Dave L. Renfro Jan 8 '19 at 17:38

I believe the claim is false for countable sets. Consider $E=\left\{\frac{1}{n}\;|\; n\in\mathbb{N}\right\}.$ Then $E'=\{0\},$ so $E'$ is closed, but not perfect since finite sets in $\mathbb R$ have no limit points.

In general, if $X$ is a separable metric space and $E\subset X,$ then $E'$ is always closed (exercise!), but $E'$ is not perfect unless $E$ is uncountable (the proof is the same for that of $\mathbb R$) or $E$ is finite, as you have observed.

For the second property, it is true that in separable metric spaces uncountable subsets have uncountably many limit points. It is also true that uncountable sets in these metric spaces contain at least one of their limit points [see this answer for a proof]. Thus, if $E\setminus E'$ were uncountable, it would contain one of its limit points, which is also a limit point of $E,$ and hence a member of $E'.$ So $(E\setminus E')\cap E'\neq \emptyset,$ a contradiction.

Thus, to summarize, all of the conclusions still hold, except that $E'$ is perfect.


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