Example 3 If X is the Cantor set, then L(X) = X.
The usual construction of the Cantor set is to start with the closed unit interval
An equivalent approach is to focus on the sequence of intervals remaining at each stage:
E | = | [0, 1] |
E0 U E1 | = | [0, 1/3] U [2/3, 1] |
E00 U E01 U E10 U E11 | = | [0, 1/9] U [2/9, 1/3] U [2/3,7/9] U [8/9,1] |
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We can think of the union of the intervals at each level as representing an approximation of the Cantor set. Say
C0 | = | E |
C1 | = | E0 U E1 |
C2 | = | E00 U E01 U E10 U E11 |
... |
Then it is easy to see
C0 contains C1 contains C2 contains ...
We see C0, C1, C2, ... is a nested sequence of sets.
The Cantor set is the intersection of all the Ci.
The set Ci consists of intervals of length 1/3i.
Note that the endpoints of every interval in every Ci belongs to all the Ci, and so belongs to the Cantor set.
Next, every point of the Cantor set is a limit point of the Cantor set.
To see this, take any point q of the Cantor set.
Then q belongs to Ci for each i. Now pick any distance d > 0.
We must show there is a point w of the Cantor set with
Two cases need to be considered.
First, suppose q is an endpoint of an interval of one of the Cj.
Then q is an endpoint of an interval in Ck for all k >= j.
No matter how small d is, there is a k >= j large enough that 1/3k < d.
Suppose q is an endpoint of an interval L in this Ck.
Then take w to be the other endpoint of L.
Certainly, w belongs to the Cantor set, and
So q is a limit point of the Cantor set.
Second, suppose q is not an endpoint of an interval in any of the Cj.
(It is not obvious that there are such points. That there are is most easily established by showing the set of all endpoints is countable, while the Cantor set is uncountable.)
Again, take k large enough that 1/3k < d.
Because q belongs to Ci for all i, q belongs to Ck, hence to an interval L of Ck.
Let w be either of the endpoints of L.
Then w is not q (the hypothesis of this second case) so 0 <
Because both w and q belong to L,
Consequently, q is a limit point of the Cantor set.
Combining these two steps, we have shown that every point of the Cantor set is a limit point of the Cantor set.
To complete this example, we must show that every limit point of the Cantor set belongs to the Cantor set.
Suppose q is a limit point of the Cantor set.
For the purpose of contradiction, suppose q does not belong to the Cantor set.
Then q must belong to
That is, q belongs to some
open interval
Let d be the smaller of the
distances
Then no point of the Cantor set lies within a distance d of q, so q cannot be a limit point of the Cantor set.
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