The Cantor Set, Baire Category, and Measure Zero in $\mathbb{R}$: Examples and Counterexamples

Let $\mathbb{Q}_{\mathbb{R}} \subseteq \mathbb{R}$ be the set of rationals (The rationals embed densely in the reals) and let $\varepsilon > 0$ be real. Then there is a sequence $(V_k)_{k \in \mathbb{N}}$ of open intervals (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with

$\mathbb{Q}_{\mathbb{R}} \subseteq \bigcup_{k \in \mathbb{N}} V_k \qquad \text{and} \qquad \sum_{k=0}^{\infty} \operatorname{length}(V_k) = \varepsilon .$

Explicitly, if $e : \mathbb{N} \to \mathbb{Q}_{\mathbb{R}}$ is a bijection, one may take $V_k := \big(e(k) - \varepsilon 2^{-k-2},\ e(k) + \varepsilon 2^{-k-2}\big)$, of length $\varepsilon 2^{-k-1}$.

This is Every at most countable subset of $\mathbb{R}$ has measure zero made concrete for the most familiar countable set, and it is the computation that makes measure zero look paradoxical: a set that meets every interval of $\mathbb{R}$ is nonetheless covered by open intervals whose lengths add up to a millionth.

• The union is a dense open set of arbitrarily small total length. Each $V_k$ is open, so $\bigcup_k V_k$ is an open set containing every rational, hence dense; and its covering intervals have total length $\varepsilon$. Iterating this over a sequence of shrinking $\varepsilon$ is exactly the construction of $\mathbb{R}$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions, where the intersection of countably many such open sets turns out to be null and residual at the same time.

• Indexing. The first interval has length $\varepsilon \cdot 2^{-1}$, not $\varepsilon$: sequences here start at $k = 0$ and the total $\varepsilon 2^{-1} \sum_{k \ge 0} 2^{-k}$ is exactly $\varepsilon$. Copying the classical $\varepsilon 2^{-k}$ from a $1$-indexed source would give total $2\varepsilon$.

• What this does not show. It does not show that the union of the $V_k$ is small: that union is an open set containing a dense set, and one may not conclude anything about its own total length from the lengths of the $V_k$, since they overlap heavily. The correct statement is about the cover, not the union, and that is why Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover) is phrased in terms of covers throughout.

By 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 $\{0,1\}^{\mathbb{N}}$ a real $x$ lies in the Cantor set $C$ exactly when

$x \;=\; \Phi(a) \;=\; \sum_{k=0}^{\infty} a_k 3^{-k-1} \qquad \text{for some sequence } a \text{ with every } a_k \in \{0,2\},$

and that sequence is then unique. The membership test is therefore: $x$ has a ternary expansion using only the digits $0$ and $2$. Six points are worked out here.

The last line is the interesting one: the digits of $1/4$ alternate for ever, so $1/4$ lies in $C$ without being an endpoint of any interval removed in the construction ($1/4$ lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it).

• The digit $1$ is what the test forbids. $1/3$ has the ternary expansion $0.1000\dots$ as well as $0.0222\dots$, and it is the second that witnesses $1/3 \in C$; the test asks for the existence of an expansion with digits in $\{0,2\}$, not for every expansion to have that form. By contrast $1/2$ is not in $C$ at all, since $C \subseteq C_1 = [0,\tfrac13] \cup [\tfrac23,1]$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) while $\tfrac13 < \tfrac12 < \tfrac23$.

• The construction and the digits match stage by stage. $C_1$ keeps the points whose first digit can be taken $0$ or $2$, $C_2$ those whose first two digits can be, and so on; that correspondence is the content of 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 $\{0,1\}^{\mathbb{N}}$ and is what makes the table computable without ever drawing the intervals.

• $1/4$ is not special. Every point of $C$ whose digit sequence is not eventually constant fails to be an endpoint, and those points are the vast majority: the eventually constant sequences are at most countable while $C$ is not (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).

Let $S$ be the Smith-Volterra-Cantor set (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals). At stage $n$ exactly $2^n$ open intervals, each of length $4^{-n-1}$, are removed, so the lengths removed at that stage total $2^{n} \cdot 4^{-n-1} = 4^{-1} \cdot 2^{-n}$ and over all stages they total

$\sum_{n=0}^{\infty} 4^{-1} \cdot 2^{-n} \;=\; 4^{-1} \cdot 2 \;=\; \tfrac12 .$

Correspondingly, no cover of $S$ by intervals has total length below $\tfrac12$: if $(a_k)$, $(b_k)$ are sequences of reals with $a_k \le b_k$, $S \subseteq \bigcup_k [a_k,b_k]$ and $\sum_{k<i}(b_k - a_k) \le M$ for every $i$, then $M \ge \tfrac12$ (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).

The two numbers are the two halves of the unit interval's length, and the second is what "the set has positive measure" means in the vocabulary available here: this library defines no outer measure, so the assertion is about covers and their total lengths, never about a number attached to $S$ itself.

• What the numbers do and do not say. "Total length of the removed intervals" is a sum of lengths of an explicit family, and "no cover below $1/2$" is a statement about all covers. Neither says that $S$ has measure $1/2$: that would require an outer measure, which is not defined at this point in the reading order. The pair of statements is nevertheless the exact content of the classical assertion.

• Why $4^{-n}$ and not $3^{-n}$. For the middle-thirds construction the removed length at stage $n$ is $2^{n}3^{-n-1}$, and $\sum_n 2^{n}3^{-n-1} = 3^{-1} \cdot 3 = 1$, so everything is removed in the sense of total length and the Cantor set is null (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). Here the removed pieces shrink faster than they multiply and only half the length goes.

• The bound $1/2$ is sharp in one direction only. The removed intervals together with a cover of $S$ must reach total length $1$, so a cover of $S$ cannot do better than $1/2$; whether total length exactly $1/2$ is approached by covers of $S$ is a question about outer measure and is not asked here.

$\mathbb{R}$ is uncountable (Finite, countably infinite, countable, uncountable), by Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets: a singleton is nowhere dense, so a listing of $\mathbb{R}$ would present $\mathbb{R}$ as a countable union of nowhere dense sets, which the Baire theorem forbids.

This is the third proof of the fact in this library. The first is Cantor's nested-interval argument of 1874 ($\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)); the second is the perfect-set theorem applied to a closed interval (Every nonempty perfect subset of $\mathbb{R}$ is uncountable); this one isolates what the first two have in common, namely completeness used through nested intervals, and packages it once.

• The proof is not circular. Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets is proved from the nested interval property and an enumeration of $\mathbb{Q}$, and it nowhere uses the uncountability of $\mathbb{R}$; nor does it use Every nonempty perfect subset of $\mathbb{R}$ is uncountable. What it shares with both is A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$, and that is the one ingredient no proof of uncountability here avoids.

• It proves more than uncountability. The same argument shows that $\mathbb{R}$ is not a countable union of nowhere dense sets, of which "not a countable union of singletons" is the weakest case. So it also shows, for instance, that $\mathbb{R}$ is not the union of countably many Cantor sets, each of which is nowhere dense (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).

• What it does not give. It gives no cardinality beyond "not at most countable", and in particular says nothing about a bijection with $\mathcal{P}(\mathbb{N})$. For the Cantor set that stronger information is available through the digit description (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 $\{0,1\}^{\mathbb{N}}$), and it is what makes FALSE: the Cantor set is countable because only countably many intervals were removed fail so badly.

Let $c$ be the Cantor function (The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval). Then

$c(x) = \tfrac12 \ \text{ for every } x \in [\tfrac13, \tfrac23], \qquad c(\tfrac19) = \tfrac14, \qquad c(\tfrac14) = \tfrac13, \qquad c(\tfrac79) = \tfrac34 .$

Each value is computed by halving the ternary digits and reading the result in base two, which is what The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval prescribes on $C$, and by the constancy across gaps of The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set off $C$.

• The staircase is visible in these five numbers. $c$ rises from $c(\tfrac19) = \tfrac14$ to $c(\tfrac13) = \tfrac12$ across the second stage, stays at $\tfrac12$ across the whole middle third, and reaches $c(\tfrac79) = \tfrac34$; between those it is constant on each removed interval (The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set). All of the rise happens on $C$, a set of measure zero (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).

• $c(1/4) = 1/3$ is the value at a non-endpoint. $1/4$ lies in $C$ and is the endpoint of no removed interval ($1/4$ lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it), so $c$ is not constant on any neighbourhood of it; the computation of step 1.2 is the only one of the five that cannot be read off a finite digit string.

• Halving digits is a bijection, not an approximation. The value $\gamma(x)$ is the exact sum of the halved-digit series, and the two series are compared term by term, never numerically; this is why $c(\tfrac13)$ and $c(\tfrac23)$ come out equal, the sequences $(0,2,2,2,\dots)$ and $(2,0,0,0,\dots)$ halving to $(0,1,1,1,\dots)$ and $(1,0,0,0,\dots)$, both summing to $\tfrac12$.

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.

Refuted claim: every nowhere dense subset of $\mathbb{R}$ has measure zero (FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero).

The witness is the Smith-Volterra-Cantor set $S$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \ge 1$, an open middle interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals): the subset of $[0,1]$ obtained by removing, at stage $n$, an open interval of length $4^{-n-1}$ from the middle of each of the $2^{n}$ intervals then present. It is nowhere dense (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$) and no cover of it by intervals has total length below $2^{-1}$, so it is not of measure zero (Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover)). This item records the witness and says what makes it work; the refutation is carried out in full in FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero and The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero.