The Cantor set contains no interval of positive length yet has no isolated point, so every connected subset of it is a single point

The Cantor set $C$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) has two properties that sound incompatible and are not:

1. it is perfect (Perfect subset of $\mathbb{R}$: closed with no isolated points): closed, and every one of its points is a limit of other points of $C$;
2. it contains no interval with two distinct endpoints, and consequently every nonempty connected subset of $C$ (Separated sets, disconnection, and connected subset of $\mathbb{R}$) is a single point.

Both are claims of The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points; this item spells out what they say together and why they do not conflict. A set can be clustered everywhere and nowhere thick: every neighbourhood of a point of $C$ contains other points of $C$, and yet no two points of $C$ are joined by a segment lying in $C$. (Not "on both sides": $C \subseteq [0,1]$ and both $0$ and $1$ lie in $C$, so the endpoints have points of $C$ approaching them from one side only. Having no isolated point is the claim, and it does not require approach from both sides.)

On the phrase "totally disconnected". That is the usual name for property 2, and no definition of it exists at this point in the reading order; the phrase appears here only as a gloss, never as the claim. What is asserted is exactly property 2 as displayed, obtained from A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval.

• The two properties pull in opposite directions and both hold. Perfectness says $C$ is nowhere sparse within itself: every neighbourhood of a point of $C$ contains infinitely many points of $C$. Property 2 says $C$ is nowhere thick in $\mathbb{R}$: it contains no segment. The Cantor set is the standard demonstration that these are independent, and it is also why "perfect" cannot be paraphrased as "contains an interval".

• Uncountability follows from the first property alone. Every nonempty perfect subset of $\mathbb{R}$ is uncountable (Every nonempty perfect subset of $\mathbb{R}$ is uncountable), so property 1 already forces $C$ to have more than countably many points, with no reference to digits. The digit route gives the same conclusion and more, an explicit bijection with $\{0,1\}^{\mathbb{N}}$ (The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$).

• The same pair of properties holds for the fat Cantor set (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero), which is also compact, perfect and nowhere dense. What differs there is only the measure, so neither property above has anything to do with total length.