Confused about the measure of the Cantor Set, and how to reconcile this with there being points not at the endpoints

Question

I know that there are points not at the endpoints of the intervals we have removed in the cantor set, such as $1\over 4$, because we can represent the numbers remaining by all ternary strings using only $0$'s or $2$'s. However, this seems to contradict the fact that we keep on removing $1\over 3$ of what we have at each stage of removing the middle third interval, so eventually we will get $\lim_{n\rightarrow \infty} (\frac 2 3)^n = 0$ left, such that nothing remains except the endpoints. How can I reconcile these 2 things, finding the correct way to think about them?

Answer 1

so eventually we will get $\lim_{n\rightarrow \infty} (\frac 2 3)^n = 0$ left, such that nothing remains except the endpoints.

This sentence is where your intuition has led you astray, and led you to leap to an unjustified conclusion. You are making a completely spurious connection between the total length of the set of remaining points (which can be defined rigorously as the infimal sum of lengths of intervals needed to cover the set), and the quality of being an endpoint.

At the risk of playing psychologist, this may be because subconsciously, you have reversed the implication $$S \text{ is the set of endpoints of disjoint intervals} \implies S \text{ has length }0\text{,}$$ which is true, to $$S \text{ has length }0 \implies S \text{ is the set of the endpoints of disjoint intervals,}$$ which is false.

The first example to correct your intuition, even though it won't directly clarify the Cantor set issue for you, should be that any countable set, even a dense one such as $\mathbb Q$, must have length $0$. To see this, simply cover the points with a sequence of intervals of length $\frac{\epsilon}{2^n}$ for arbitrarily small $\epsilon$.

But certainly $\mathbb Q$ is not the set of endpoints for a union of disjoint intervals.

Now that you realize that length zero doesn't have anything to do with endpoints, it should be easier to accept this Cantor example, which you can use to further kill off any intuition you might be tempted to have that $0$ length even implies countable.