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Let $K$ be a non-empty compact perfect set of reals.

Question. Is it possible that $K$ is not totally disconnected and does not contain regular closed subsets?

Intuitively the answer should be negative (removing each time the closed interval subsets).

It is a variant of this question, where I was too late to make the necessary edits to avoid the easy answer.

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    $\begingroup$ What do you mean by regular closed subsets? $\endgroup$ – Fimpellizieri Aug 24 '17 at 19:45
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    $\begingroup$ A set is regular closed if it is equal to the closure of its interior (it was recalled in the linked thread) $\endgroup$ – Paolo Leonetti Aug 24 '17 at 19:49
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If $K$ is not totally disconnected, it contains a non-trivial connected subset $I$. The only connected subsets of $\mathbb R$ are intervals, so by considering a smaller interval if necessary we have $[a,b]\subset K$ for some $a<b$. This is a regular closed set. The assumptions of compactness and perfectness seem unnecessary.

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Every subspace contains regular closed subsets: just take the closure of any open set. These are regular closed in the subspace topology.

If $K$ is perfect in the reals, then it either contains an interval or it is totally disconnected. If the former holds, this closed interval is regular closed (in the reals as well) and if the latter, no closed subset of $K$ can be perfect in the reals.

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  • $\begingroup$ The proposer did not state whether regular closed meant so in $\mathbb R$ or in $K.$...............+1 $\endgroup$ – DanielWainfleet Aug 25 '17 at 11:35

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