TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Distinct points of a metric space have disjoint balls around them

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 $p, q \in X$ with $p \neq q$ . Put $r : = d (p, q) / 2$ . Then $r > 0$ and

$B (p, r) \cap B (q, r) = \emptyset .$

Both sets are open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed) and contain $p$ respectively $q$ (Open ball, closed ball and sphere in a metric space), so every metric space is Hausdorff: distinct points are separated by disjoint open sets (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).

Facts & Assumptions

Given: A metric space $(X, d)$ and points $p, q \in X$ with $p \neq q$ ; write $c : = d (p, q)$ .

[L1]

Separation (M1), symmetry (M2) and the triangle inequality (M3) of a metric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); and $d \geq 0$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L2]

Halving. For a real $c > 0$ put $2 : = 1 + 1$ and $c / 2 : = c \cdot 2^{- 1}$ . Then $2 > 0$ , so $2 \neq 0$ and $2^{- 1} > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field); hence $c / 2 > 0$ (Sign rules for products and monotonicity of multiplication); and $c / 2 + c / 2 = c (2 \cdot 2^{- 1}) = c$ (Field).

[L3]

Adding two strict inequalities: $a < b$ and $a^{'} < b^{'}$ give $a + a^{'} < b + b^{'}$ (Order is preserved by adding a constant and by adding inequalities).

[L4]

Trichotomy of the order of $R$ : $a < a$ is impossible, and $a \neq 0$ together with $a \geq 0$ gives $a > 0$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

Membership in a ball: $z \in B (u, t)$ means $d (u, z) < t$ ; balls are open and contain their centres (Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

Proof

technique · direct

1.1

Since $p \neq q$ , axiom (M1) gives $c = d (p, q) \neq 0$ , and $c \geq 0$ , so $c > 0$ by trichotomy; hence $r : = c / 2$ is a positive real with $r + r = c$ .

givenL1L2L4

2.1

Suppose some $z \in X$ lay in both $B (p, r)$ and $B (q, r)$ , that is $d (p, z) < r$ and $d (q, z) < r$ ; then symmetry and the triangle inequality give $c = d (p, q) \leq d (p, z) + d (z, q) = d (p, z) + d (q, z) < r + r = c$ , so $c < c$ , which trichotomy forbids.

step 1.1L1L3L4L5

3.1

No such $z$ exists, so $B (p, r) \cap B (q, r) = \emptyset$ ; both sets are open and contain $p$ respectively $q$ , so distinct points of $(X, d)$ are separated by disjoint open sets.

step 1.1step 2.1L5∎

Remarks

This is a strengthening of uniqueness of limits. A sequence converging to two distinct points would eventually be inside both $B (p, r)$ and $B (q, r)$ , which the theorem forbids; that is a second route to A sequence in a metric space has at most one limit, and the two proofs use the same halving.
Separation (M1) is what the proof spends. A pseudometric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) with $d (p, q) = 0$ for distinct $p, q$ gives $r = 0$ at step 1.1 and there is no ball to speak of; such a space is not Hausdorff, and no argument repairs that.
The radius $d (p, q) / 2$ is not the only choice, and it is not optimal in every space: in an ultrametric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) the balls $B (p, c)$ and $B (q, c)$ are already disjoint, because a common point would force $d (p, q) \leq \max {d (p, z), d (z, q)} < c$ .

Depends on

Used by

Dependency tree · two levels

20 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

Hausdorff space (Wikipedia) (standard reference, not scraped)
Metric space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §17 (standard reference, not scraped)