TITLE: Deleting intervals from $(0,1):$ what is the resulting set?
QUESTION [6 upvotes]: Take the open interval $(0,1)$.

Split into thirds and consider only the open intervals at each end. We are left with:
$$\big(0,\tfrac13\big)\cup \big(\tfrac23,1\big)$$ For each of these:

Split into quarters and consider only the open intervals at each end. We are left with: $$\big(0,\tfrac{1}{12}\big)\cup \big(\tfrac{1}{4},\tfrac{1}{3}\big)\cup \big(\tfrac23,\tfrac34\big)\cup \big(\tfrac{11}{12},1\big)$$
For each of these:

Split into fifths $...$


$$\text{etc}.$$
Continuing this process indefinitely (a countably infinite amount of times), what are we left with?
Initially I would have said nothing, but having read and asked about similar Cantor set like things, I am not so sure...

REPLY [3 votes]: You don't get nothing -- every infinite sequence of "left" and "right" that doesn't eventually end in all "left"s or all "right"s will map to a sequence of nested open intervals where both right and left endpoints move towards each other. They therefore have a point in common that never gets removed.
In fact, if you use closed intervals instead of open ones, you get something that is homeomorphic to the usual middle-thirds Cantor set, so with open intervals the result is homeomorphic to the middle-thirds Cantor set without its interval endpoints.
It can be viewed as a "thin" analogue of a fat Cantor set; how large a fraction of each interval you're removing at each step doesn't change the topology of the limit set (though it can change its measure, as the fat Cantor set shows).