nonempty perfect subset of that contains no rational number, a
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 for some irrational numbers and , with . Let be an enumeration of the rational numbers in . For each , we will define an open interval and delete it.
Let and be two irrational numbers such that . Define .
Having defined , and , let’s define and :
If then there exists an such that . Let and .
Otherwise let and be two irrational numbers such that , and which satisfy:
Now define . Note that by our choice of and any of the previous end points are not removed from .
Now let us see that does not have any isolated points. Let , and be given. If for any , choose a rational number such that . Then and since we must have , which means . Since we know that , this shows that is a limit point. Otherwise, if for some , then choose a such that . Similarly, and it follows that . We have shown that any point of is a limit point, hence is perfect.
|Title||nonempty perfect subset of that contains no rational number, a|
|Date of creation||2013-03-22 15:26:10|
|Last modified on||2013-03-22 15:26:10|
|Last modified by||Gorkem (3644)|