LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $x \in X$ . For a natural $n \geq 1$ write $1 / n$ for the inverse of the canonical natural $n \cdot 1_{R}$ , a positive real, and put

$β_{n} : = B (x, 1 / n), B_{x} : = {β_{n} : n \in N, n \geq 1} .$

Then:

$B_{x}$ is at most countable (Finite, countably infinite, countable, uncountable).
Every $β_{n}$ is an open subset of $X$ containing $x$ .
For every open $U \subseteq X$ with $x \in U$ there is $n \geq 1$ with $β_{n} \subseteq U$ .

The two names used in the title are introduced by this statement, not cited from elsewhere. A family of open sets each containing $x$ , such that every open set containing $x$ contains a member of the family, is a neighbourhood base at $x$ ; a space in which every point has an at most countable neighbourhood base is first countable. Claims 1 to 3 say that $B_{x}$ is an at most countable neighbourhood base at $x$ , so every metric space is first countable.

Facts & Assumptions

Given: A metric space $(X, d)$ , a point $x \in X$ , and for each natural $n \geq 1$ the ball $β_{n} = B (x, 1 / n)$ .

[L1]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing), hence $n \cdot 1_{R}$ is invertible with $1 / n > 0$ (Inverses of positives are positive, and reciprocation reverses order); and $N$ contains $0$ , so $j \mapsto j + 1$ runs over exactly the naturals $\geq 1$ as $j$ runs over $N$ (The natural numbers $N$ (von Neumann)).

[L2]

Balls are open and every ball contains its centre (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space); and $B (x, s) \subseteq B (x, t)$ whenever $0 < s \leq t$ (Open ball, closed ball and sphere in a metric space).

[L3]

Open sets: $U$ is open when every point of $U$ has a ball around it inside $U$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L4]

Reciprocal Archimedean property: for every real $r > 0$ there is a natural $n \geq 1$ with $1 / n < r$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L5]

Countability: a nonempty set admitting a surjection from $N$ is at most countable (A nonempty set is at most countable iff it is a surjective image of $N$ , Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

Proof

technique · direct

1.1

For every natural $n \geq 1$ the real $1 / n$ is defined and positive, so $β_{n}$ is a legitimate ball of positive radius.

1.2

Let $U$ be open with $x \in U$ , and fix a real $r > 0$ with $B (x, r) \subseteq U$ ; then fix a natural $n \geq 1$ with $1 / n < r$ .

L3L4choose

2.1

Each $β_{n}$ is open and contains $x$ , which is claim 2.

step 1.1L2

2.2

The map $s : N \to B_{x}$ given by $s (j) : = β_{j + 1}$ is well defined by step 1.1 and is surjective, because every member of $B_{x}$ is $β_{n}$ for some $n \geq 1$ and $n = j + 1$ for the natural $j$ with $j + 1 = n$ ; moreover $B_{x}$ is nonempty, containing $β_{1}$ .

step 1.1L1

2.3

By step 1.2 and monotonicity of balls in the radius, $β_{n} = B (x, 1 / n) \subseteq B (x, r) \subseteq U$ , which is claim 3.

step 1.2L2

3.1

By [L5] applied to the surjection of step 2.2, the nonempty set $B_{x}$ is at most countable, which is claim 1.

step 2.2L5

4.1

Claims 1, 2 and 3 hold by steps 3.1, 2.1 and 2.3, so $B_{x}$ is an at most countable neighbourhood base at $x$ and $(X, d)$ is first countable.

step 2.1step 2.3step 3.1∎

Remarks

The family can be finite, and that is not a defect. In a discrete metric space $1 / n \leq 1$ for every $n \geq 1$ , so every $β_{n}$ is the single point ${x}$ and $B_{x}$ is a one-element family. "At most countable" in this library includes finite (Finite, countably infinite, countable, uncountable), which is exactly why claim 1 is stated in that form and not as "countably infinite".
Only the reciprocal form of the Archimedean property is used, in step 1.2, and it is the form recorded as For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ precisely so that the inversion never has to be redone inside a proof.
This is what makes sequences sufficient in metric spaces. First countability is the hypothesis under which closure can be described by sequences rather than by nets, and it is used in exactly that way by A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed.

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

First-countable space (Wikipedia) (standard reference, not scraped)
Neighbourhood system (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §30 (standard reference, not scraped)

The balls B(x,1/n)B(x,1/n), n≥1n≥1, form a countable neighbourhood base at xx, so every metric space is first countable