# Non empty Perfect set with only irrational numbers [duplicate]

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I encountered this question in Abbott's Understanding Analysis. The problem asks to construct a nonempty perfect set with no rationals. It starts with enumerating the rationals $$\mathbb{Q}=\{r_1,r_2,...\}$$. At first we start with an open set $$O:=\bigcup_1^\infty V_{\epsilon_n}(r_n)$$ where $$\epsilon_n:=1/2^n$$ and $$V_{\epsilon_n}(r_n)$$ is open neighborhood of $$r_n$$ with radius $$\epsilon_n$$. Then $$F:=O^c$$ is obviously closed and nonempty (since lengths of the open intervals add up to 2) and $$F$$ obviously contains only irrationals also $$F$$ does not contain any open intervals since it does not contain any rational so $$F$$ is totally disconnected as well. But it is not possible to know that $$F$$ is perfect, since it may have isolated points. How do I modify the above construction to make $$F$$ as a perfect set??

Edit: I have seen similar questions here but with different constructions of different sets. Here the question asks for a very specific modification of a given construction.

## marked as duplicate by Eric Towers, Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 21 at 8:01

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• Yes I have seen similar questions. But the context is different. There are indeed many proofs possible using many fancy constructions. This problem asks a modification of a very specific construction. – Arpan Das Jun 21 at 7:50
• Infact one can easily construct a cantor set like thing by avoiding the $n-th$ rational at the $n-th$ stage of construction by removing large enough open sets from the $n-1$ th stage. – Arpan Das Jun 21 at 7:53
• Every uncountable closed set contains a perfect subset. That is, in effect, Cantor's first step towards proving CH. – Asaf Karagila Jun 21 at 8:04
• Every uncountable Borel set contains a subspace homeomorphic to the Cantor set. The irrationals are a Borel set. – Henno Brandsma Jun 21 at 9:36