ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Baire category gives a third proof that $\mathbb{R}$ is uncountable

Example

$\mathbb{R}$ 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.

Facts & Assumptions

Given: The complete ordered field $\mathbb{R}$ .

[L1]

If $(A_n)_{n \in \mathbb{N}}$ is a sequence of nowhere dense subsets of $\mathbb{R}$ then $\bigcup_n A_n \ne \mathbb{R}$ (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).

[L2]

A closed set is nowhere dense exactly when its interior is empty (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$ , Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points).

[L3]

$U$ is open when every point of it has a neighbourhood inside it, $F$ is closed when its complement is open, and $N_\varepsilon(x) = (x - \varepsilon, x + \varepsilon)$ contains $x$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

A nonempty at most countable set admits a surjection from $\mathbb{N}$ , and uncountable means not at most countable (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$ , Finite, countably infinite, countable, uncountable, Injection, surjection, bijection).

[L5]

$\mathbb{R}$ is uncountable, by Cantor's nested-interval argument ( $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)).

[L6]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ and $0 < \varepsilon \cdot 2^{-1} < \varepsilon$ for $\varepsilon > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

For $c \in \mathbb{R}$ the singleton $\{c\}$ is nowhere dense: it is closed, since $x \ne c$ gives $N_{|x-c|}(x) \subseteq \mathbb{R} \setminus \{c\}$ by [L3]; and its interior is empty, since for every real $\varepsilon > 0$ the point $c + \varepsilon \cdot 2^{-1}$ lies in $N_\varepsilon(c)$ and differs from $c$ by [L6], so no neighbourhood of $c$ is contained in $\{c\}$ . By [L2] it is nowhere dense.

L2L3L6

2.1

Let $s : \mathbb{N} \to \mathbb{R}$ be any function. The sets $A_n := \{s(n)\}$ are nowhere dense by step 1.1, so $\bigcup_n A_n \ne \mathbb{R}$ by [L1]; but $\bigcup_n A_n$ is exactly the image of $s$ , so $s$ is not surjective.

step 1.1L1

3.1

Hence there is no surjection $\mathbb{N} \to \mathbb{R}$ . Since $\mathbb{R}$ is nonempty, [L4] gives that $\mathbb{R}$ is not at most countable, that is, $\mathbb{R}$ is uncountable, which is [L5] reproved along an independent route.

step 2.1L4L5∎

Remarks

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 107 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Baire category theorem (Wikipedia) (standard reference, not scraped)
Cantor's first set theory article (Wikipedia) (standard reference, not scraped)
Baire theorem (Encyclopedia of Mathematics) (standard reference, not scraped)
E. Zakon, Mathematical Analysis, §6.8: Baire Categories (standard reference, not scraped)

Baire category gives a third proof that R\mathbb{R}R is uncountable