TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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The closure of a nonempty $A$ is $\{x : d(x,A) = 0\}$ , equals $A$ together with its limit points, and is the smallest closed superset

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 $A \subseteq X$ , with closure, derived set and limit points as in Interior, closure, boundary, limit point, isolated point and dense subset of a metric space. Then:

If $A \ne \emptyset$ , then $\overline{A} = \{\, x \in X : d(x,A) = 0 \,\}$ , where $d(x,A)$ is the distance from a point to a nonempty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).
$\overline{A} = A \cup A'$ .
$\overline{A}$ is closed, contains $A$ , and is contained in every closed $F \subseteq X$ with $A \subseteq F$ . So $\overline{A}$ is the smallest closed superset of $A$ , and $A$ is closed if and only if $A = \overline{A}$ .

Claims 2 and 3 hold for every $A$ , the empty set included: $\overline{\emptyset}$ is empty because no ball meets $\emptyset$ , and $\emptyset$ is closed because $X$ is open (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). Claim 1 carries the hypothesis $A \ne \emptyset$ because $d(x,A)$ is defined only for nonempty $A$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Facts & Assumptions

Given: A metric space $(X,d)$ , a subset $A \subseteq X$ , a point $x \in X$ , and a closed set $F \subseteq X$ with $A \subseteq F$ ; when $A \ne \emptyset$ , the set $E(x) := \{\, d(x,a) : a \in A \,\}$ , whose infimum is $d(x,A)$ .

[A1]

Closure and derived set: $x \in \overline{A}$ means $B(x,r) \cap A \ne \emptyset$ for every $r > 0$ ; $x \in A'$ means $B(x,r) \cap (A \setminus \{x\}) \ne \emptyset$ for every $r > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[A2]

Open and closed: $U$ is open when every point of $U$ has a ball around it inside $U$ ; $F$ is closed when $X \setminus F$ is open (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).

[L1]

For nonempty $A$ , the set $E(x)$ is nonempty and bounded below by $0$ , so $d(x,A) = \inf E(x)$ exists and is a lower bound of $E(x)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

[L2]

Epsilon characterisation of the infimum: for a nonempty $S \subseteq \mathbb{R}$ bounded below and a lower bound $\ell$ of $S$ , one has $\ell = \inf S$ if and only if for every $\varepsilon > 0$ there is $s \in S$ with $s < \ell + \varepsilon$ (Epsilon characterisation of the infimum).

[L3]

Balls are open, so a point of a ball has a ball around it inside that ball (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).

[L4]

Membership in a ball: $a \in B(x,r)$ means $d(x,a) < r$ , and $x \in B(x,r)$ always (Open ball, closed ball and sphere in a metric space); trichotomy of the order of $\mathbb{R}$ (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct

1.1

Suppose $A \ne \emptyset$ and $x \in \overline{A}$ , and let $\varepsilon > 0$ be arbitrary; then $B(x,\varepsilon) \cap A \ne \emptyset$ , so there is $a \in A$ with $d(x,a) < \varepsilon = 0 + \varepsilon$ , and $0$ is a lower bound of $E(x)$ , so $d(x,A) = 0$ by the epsilon characterisation.

A1L1L2L4

1.2

Conversely suppose $A \ne \emptyset$ and $d(x,A) = 0$ , and let $r > 0$ be arbitrary; the epsilon characterisation supplies $a \in A$ with $d(x,a) < 0 + r = r$ , that is $a \in B(x,r) \cap A$ , so $x \in \overline{A}$ .

A1L1L2L4

1.3

$A \subseteq \overline{A}$ and $A' \subseteq \overline{A}$ : a point $a \in A$ lies in $B(a,r) \cap A$ for every $r$ , and a ball meeting $A \setminus \{x\}$ meets $A$ .

A1L4

1.4

If $x \in \overline{A}$ and $x \notin A$ , then for every $r$ the nonempty set $B(x,r) \cap A$ equals $B(x,r) \cap (A \setminus \{x\})$ , since $x$ is not a member of $A$ ; hence $x \in A'$ .

1.5

$\overline{A}$ is closed: let $x \in X \setminus \overline{A}$ and fix $r$ with $B(x,r) \cap A = \emptyset$ ; for $y \in B(x,r)$ there is $s$ with $B(y,s) \subseteq B(x,r)$ , so $B(y,s) \cap A = \emptyset$ and $y \notin \overline{A}$ , whence $B(x,r) \subseteq X \setminus \overline{A}$ and $X \setminus \overline{A}$ is open.

A1A2L3

1.6

$\overline{A} \subseteq F$ for every closed $F \supseteq A$ : if $x \in \overline{A}$ had $x \notin F$ , then $X \setminus F$ open would give $r$ with $B(x,r) \subseteq X \setminus F \subseteq X \setminus A$ , so $B(x,r) \cap A = \emptyset$ , contradicting $x \in \overline{A}$ .

A1A2given

2.1

Claim 1 follows: by step 1.1 every adherent point of a nonempty $A$ satisfies $d(x,A) = 0$ , and by step 1.2 every $x$ with $d(x,A) = 0$ is adherent.

step 1.1step 1.2

2.2

Claim 2 follows: $A \cup A' \subseteq \overline{A}$ by step 1.3, and $\overline{A} \subseteq A \cup A'$ by step 1.4, since a point of $\overline{A}$ either lies in $A$ or, not lying in $A$ , lies in $A'$ .

step 1.3step 1.4

2.3

Claim 3 follows: $\overline{A}$ is closed by step 1.5, contains $A$ by step 1.3, and sits inside every closed superset of $A$ by step 1.6; in particular if $A$ is closed then $\overline{A} \subseteq A \subseteq \overline{A}$ , so $A = \overline{A}$ , and conversely if $A = \overline{A}$ then $A$ is closed.

step 1.3step 1.5step 1.6

3.1

Claims 1, 2 and 3 are therefore all established.

step 2.1step 2.2step 2.3∎

Remarks

Claim 1 is where the infimum does the work. Reading it right to left, $d(x,A) = 0$ says that $A$ has points arbitrarily close to $x$ without saying that any of them is $x$ ; reading it left to right, adherence says the same thing in the language of balls. The equivalence is exactly the epsilon characterisation of the infimum (Epsilon characterisation of the infimum) with the lower bound $0$ .
The distance function is $1$ -Lipschitz ( $|d(x,A) - d(y,A)| \le d(x,y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz), so claim 1 exhibits $\overline{A}$ as the zero set of a function that does not increase distances. That is not used above and is recorded only as orientation.
Claim 3 is the form that transfers to general topology, where no metric is available and the closure is defined outright as the intersection of all closed supersets. Claim 1 is the specifically metric statement, and claim 2 sits between them.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 11 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

Closure (topology) (Wikipedia) (standard reference, not scraped)
Hausdorff distance (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

The closure of a nonempty AAA is {x:d(x,A)=0}\{x : d(x,A) = 0\}{x:d(x,A)=0}, equals AAA together with its limit points, and is the smallest closed superset