nonempty perfect subset of R that
contains no rational number, a∗
Gorkem†
2013-03-21 19:32:02
We will construct a nonempty perfect set contained in R that contains no
rational number.
We will begin with a closed interval, 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 :
n
[
If qn+1 ∈
(ak , bk ) then there exists an i ≤ n such that qn+1 ∈ (ai , bi ). Let
k=1
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:
qn+1 − an+1 <
min
{|qn+1 − bi |}
min
{|ai − qn+1 |}.
i=0,1,2,...,n
and
bn+1 − qn+1 <
i=0,1,2,...,n
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 =
En . E is clearly nonempty, does not contain any rational
n=1
number, and also it is compact, being an intersection of compact sets.
∗ hNonemptyPerfectSubsetOfmathbbRThatContainsNoRationalNumberAi
created: h201303-21i by: hGorkemi version: h37283i Privacy setting: h1i hExamplei h54A99i
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Now let us see that E does not have any isolated points. Let x ∈ E, and
> 0 be given. If x 6= aj for any j ∈ {0, 1, 2, . . .}, choose a rational number qk
such that x < qk < x + . Then qk ∈ (ak , bk ) and since x ∈ E we must have
x < ak , which means ak ∈ (x, x + ). Since we know that ak ∈ E, this shows
that x is a limit point. Otherwise, if x = aj for some j, then choose a qk such
that x − < qk < x. Similarly, qk ∈ (ak , bk ) and it follows that bk ∈ (x − , x).
We have shown that any point of E is a limit point, hence E is perfect.
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