nonempty perfect subset of that contains no rational number, a

We will construct a nonempty perfect setMathworldPlanetmath contained in that contains no rational number.

We will begin with a closed intervalMathworldPlanetmath, and then, imitating the construction of Cantor set, we will inductively delete each rational number in it together with an open interval. We will do it in such a way that the end points of the open intervals will never be deleted afterwards.

Let E0=[b0,a0] for some irrational numbers a0 and b0, with b0<a0 . Let {q1,q2,q3,} be an enumeration of the rational numbers in [b0,a0]. For each qi, we will define an open interval (ai,bi) and delete it.

Let a1 and b1 be two irrational numbers such that b0<a1<q1<b1<a0. Define E1=E0\(a1,b1).

Having defined E1,E2,,En, a1,a2,,an and b1,b2,,bn, let’s define an+1 and bn+1:

If qn+1k=1n(ak,bk) then there exists an in such that qn+1(ai,bi). Let an+1=ai and bn+1=bi.

Otherwise let an+1 and bn+1 be two irrational numbers such that b0<an+1<qn+1<bn+1<a0, and which satisfy:




Now define En+1=En\(an+1,bn+1). Note that by our choice of an+1 and bn+1 any of the previous end points are not removed from En.

Let E=n=1En. E is clearly nonempty, does not contain any rational number, and also it is compactPlanetmathPlanetmath, being an intersectionMathworldPlanetmath of compact sets.

Now let us see that E does not have any isolated points. Let xE, and ϵ>0 be given. If xaj for any j{0,1,2,}, choose a rational number qk such that x<qk<x+ϵ. Then qk(ak,bk) and since xE we must have x<ak, which means ak(x,x+ϵ). Since we know that akE, this shows that x is a limit pointMathworldPlanetmathPlanetmath. Otherwise, if x=aj for some j, then choose a qk such that x-ϵ<qk<x. Similarly, qk(ak,bk) and it follows that bk(x-ϵ,x). We have shown that any point of E is a limit point, hence E is perfect.

Title nonempty perfect subset of that contains no rational number, a
Canonical name NonemptyPerfectSubsetOfmathbbRThatContainsNoRationalNumberA
Date of creation 2013-03-22 15:26:10
Last modified on 2013-03-22 15:26:10
Owner Gorkem (3644)
Last modified by Gorkem (3644)
Numerical id 18
Author Gorkem (3644)
Entry type Example
Classification msc 54A99