Before we get too carried away, though, it might help matters if we first give a formal definition of the Cantor set. Let's begin by labeling the sets at "finite levels" in our procedure.
And so on... With this notation, the Cantor set is formally defined to be the intersection of all these sets:
Notice that we have constructed a decreasing sequence of nested sets. That is,
and so truly is the "limit" of the 's.
To see that we haven't just given an elaborate definition for an empty set, let's next recall (a portion of) our initial construction, this time labeling the endpoints.
Since we always remove points from the middle of an interval, notice that the endpoints 0 and 1 remain after the first "deletion". Both 0 and 1 remain after the second and third deletions, too. In fact, they remain after every subsequent deletion. Thus, the limiting set contains at least the two endpoints 0 and 1.
But now we just apply the same reasoning to the points 1/3 and 2/3. Notice that they, too, remain in our set after any deletion. The same is true of 1/9, 2/9, 7/9, and 8/9. In short, will necessarily contain the endpoints of any "discarded middle thirds" intervals; namely,
We'll call these the endpoints of . It's clear that there are infinitely many endpoints of (since there are infinitely many discarded middle thirds intervals, and each generates two such endpoints). In particular, evidently contains infinitely many points!
By taking just a slightly different viewpoint, we can say even more: The Cantor set actually contains uncountably many points. As we'll see, the Cantor set contains "as many" points as the interval [0,1]. In particular, the Cantor set evidently contains many more points than the mere handful of endpoints we've displayed above!
To find out why, proceed to still more on the Cantor set. Or go back to my homepage (just as the story was getting good, too).