We next set up a correspondence, or "matching", between the points in the Cantor set and the points in the interval [0,1]. In order to do this, we'll take a fresh look at our construction.

Instead of removing middle thirds, imagine our construction as a process of retaining the left and right thirds of each interval, labeling our choices as we go along.

What we arrive at is a strange "stairstep", with the steps labeled L and R.

Somewhere, far below this diagram, lie the points of . In order to reach a given point in , we would follow a "path" down this stairstep, choosing left or right at each step, always careful to choose a subinterval of our current "step". (You're not allowed to jump across a gap in the current step; you just step down, choosing either the left or right third of the current step.)

Any
given sequence of choices

necessarily determines a single point in , since the corresponding interval "steps" are nested and their lengths decrease to 0. That is, the intersection of the sequence of intervals in such a path will consist of a single point in . In other words, the left and right endpoints of our sequence of "steps" will converge to a common value.

Conversely,
each point in
uniquely determines the path, or sequence of L's and R's,
that leads to it. Indeed, given a point in
,
at the "bottom" of our stairstep, there can be but one
sequence of interval "steps" that contain it. Why?
Because, at each horizontal level of our picture,
the L- and R-intervals are **disjoint**.
A given point in
can be in only one, unique subinterval at each level.

The conclusion to be drawn from all of this is that we have a matching, or correspondence between the points of the Cantor set and the collection of all sequences of L's and R's.

This
means that the Cantor set is HUGE. To see why, proceed to
**even more on the Cantor set**.
Or, quit now and be the only kid
on the block who doesn't know the secret!

Neal Carothers - carother@bgnet.bgsu.edu