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This has already been asked several times and well answered here and here so I am asking if my approach is correct or not. My question is as follows

Suppose $E$ is a set in $[0, 1]$ whose decimal expansion consists only of $\{4, 7\}$. It can be shown that $E$ is perfect set. Is $E \setminus \Bbb Q$ perfect?

I cannot find any isolated point but I am confused about $E\setminus \Bbb Q$ being closed.

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  • $\begingroup$ Let $(X,d)$ be a complete metric space. (1). If $X$ is non-empty and has no isolated points then $X$ has a subspace $D$ which is homeomorphic to the Cantor set. So $D$ is a perfect set. (2). If $Y$ is a subspace of $X$ then the subspace topology on $Y$ can be generated by a complete metric $e$ on $Y$ iff $Y$ is a $G_{\delta}$ subset of $X.$.... By (2), $\Bbb R$ \ $\Bbb Q$ is completely metrizable. So by (1), $\Bbb R$ \ $\Bbb Q$ has a perfect subset. $\endgroup$ – DanielWainfleet Jul 17 '18 at 17:40
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No, $E\setminus\mathbb{Q}$ is not perfect. For example, consider the sequence:

  • $0.47447444744447444447...$
  • $0.447447444744447444447...$
  • $0.4447447444744447444447...$
  • $0.44447447444744447444447...$
  • $0.444447447444744447444447...$
  • $0.4444447447444744447444447...$

  • ...

Each number in the sequence is in $E\setminus\mathbb{Q}$, but the limit of the sequence is $0.44444....$ which is rational. So $E\setminus\mathbb{Q}$ is not closed.

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  • $\begingroup$ I see ... thank you :) $\endgroup$ – Hash Nuke Jul 17 '18 at 14:38

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