Are there any significant properties of that quotient space of $[0,1]$ by the Cantor set?

Question

The Cantor set has two interesting properties of being nowhere dense and uncountably infinite. I was curious what sort of quotient space is made when you take $[0,1]$ by the Cantor set. In other words, gluing all Cantor points to a single point.

Answer 1

Since $[0,1]$ is compact and the Cantor set is closed, the quotient space will also be compact.

Note: I'll denote the single point of the image of the Cantor set under the canonical map as the Cantor point. I'll also denote the quotient space as $X$.

Now consider the open cover of the quotient space that consists of an arbitrary open neighbourhood of the Cantor point, and the individual images of all the open intervals taken away from $[0,1]$ to obtain the Cantor set. This clearly is an open cover of $X$, and since $X$ is compact, it must have a finite subcover.

Now obviously the intervals don't cover the Cantor point, therefore any subcover must contain the open neighbourhood of the Cantor point. But then, the finite subcover only contains a finite subset of the open intervals, therefore the open neighbourhood of the Cantor point covers all but finitely many intervals.

Note that for any point on one of the intervals, there exists a neighbourhood consisting only of points of that interval, and the subspace topology of a single interval together with the Cantor point is easily seen to be that of a circle.

Therefore the quotient space is like the Hawaiian earring, but with uncountably many circles.

Note that since the complement of the Cantor point is open, this set is the one-point compactification of $[0,1]\setminus C$.