Given an uncountable set $X\subset [0,1]$ it is easy to write $X$ as a disjoint union of a perfect set $P$ (perfect in the subspace $X$) and an at most countable set $C$: just take $P$ as the set of condensation points of $X$ and $C$ as its complement in $X$. (we consider $X$ equipped with the subspace topology.)

I want to find an open set in $X$ which is dense-in-itself.

Is this possible? An open set of a dense-in-itself/perfect space is itself dense-in-itself, but from this point onwards I am most unsure how to proceed correctly.

Any comments will be appreciated. Thank you for your help.

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  • $\begingroup$ The statement in the first line is false. For instance, a Bernstein set has the cardinality of the continuum, but it contains no perfect set. $\endgroup$ – Pedro Sánchez Terraf May 20 '15 at 11:20
  • $\begingroup$ I have expressed myself without enough clarity (thus the edit). I meant a perfect set, when considered in the subspace topology. $\endgroup$ – André Guerra May 20 '15 at 18:02

It seems the following.

I want to find an open set in $X$ which is dense-in-itself. Is this possible?

It is not always possible. Let $\{a_n\}$ be an enumeration of all rational points of the unit segment $[0,1]$. Let $$X=\{(1/n, a_n):n\in\Bbb N\}\cup (\{0\}\times ([0,1]\setminus\Bbb Q)) \subset\Bbb R^2.$$ Then $X$ is an uncountable zero-dimensional metric space with a dense set of isolated points. But $X$ belongs to $\Bbb R^2$, not to $\Bbb R$. To solve this little problem we recall that each zero-dimensional second countable space can be embedded into Cantor set, which can be embedded into the unit segment.

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(cited from “General Topology” by Ryszard Engelking. )

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  • $\begingroup$ As your answer is written, $X=\emptyset$ - I'm not sure what you meant. $\endgroup$ – André Guerra May 20 '15 at 18:01
  • $\begingroup$ @user241898 Thanks. Corrected. $\endgroup$ – Alex Ravsky May 20 '15 at 19:07
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    $\begingroup$ Nice example. Thank you. $\endgroup$ – André Guerra May 22 '15 at 18:49

This is an easier version of Alex Ravsky’s idea.

Let $C$ be the usual middle-thirds Cantor set, let $D$ be the set of midpoints of the open intervals deleted in the construction of $C$, and let $X=C\cup D$; then $D$ is a dense set of isolated points in $X$, so $X$ has no dense-in-itself open subset.

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  • $\begingroup$ This is a very simple counterexample - thank you. $\endgroup$ – André Guerra May 24 '15 at 15:43
  • $\begingroup$ @André: You’re welcome. $\endgroup$ – Brian M. Scott May 24 '15 at 21:33

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