LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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A convex set and its closure have the same interior and boundary

Statement

Assume the Axiom of Choice (The Axiom of Choice) and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $C \subseteq R^{n}$ be nonempty and convex. The closure $\overline{C}$ is convex, $int (C) = int (\overline{C})$ , and $\partial C = \partial \overline{C}$ .

Facts & Assumptions

Given: The choice principles and the Euclidean topology and inner product in the Statement The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ . Relative interior is taken inside the affine hull of the set.

[A1]

The Axiom of Choice says that every family of nonempty sets has a choice function (The Axiom of Choice).

[A2]

The Axiom of Countable Choice says that every family of nonempty sets indexed by $N$ has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A subset $U \subseteq R^{m}$ is convex when every $(1 - t) x + t y$ , for $x, y \in U$ and $t \in [0, 1]$ , belongs to $U$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L1]

Under ${A C}_{ω}$ , a point lies in the closure of a subset of a metric space exactly when it is the limit of a sequence from that subset (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).

[L2]

The span of a set is the smallest linear subspace containing it (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[L3]

Assuming AC, every spanning set of a finite-dimensional vector space contains a basis of that space (Every spanning subset of a vector space contains a basis).

[L4]

Every linear subspace of a finite-dimensional vector space is finite-dimensional, of dimension at most that of the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

[L5]

For $d \geq 1$ , every norm $N$ on $R^{d}$ admits positive constants $c, C$ with $c ∥ x ∥_{2} \leq N (x) \leq C ∥ x ∥_{2}$ for every $x$ (For $n \geq 1$ all norms on $R^{n}$ are equivalent).

[L6]

For $d \geq 1$ , Euclidean space $R^{d}$ is complete ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ).

[L7]

The boundary of $A$ is $\partial A = \overline{A} ∖ int (A)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[L8]

The closure of $A$ is the smallest closed superset of $A$ , and $A$ is closed exactly when $A = \overline{A}$ (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).

Proof

technique · direct

1.1

Using [A2], paired sequences from $C$ and [F1] show by [L1] that $\overline{C}$ is convex. Fix $c_{0} \in C$ and put $W = span (C - c_{0})$ and $A = c_{0} + W$ , the affine hull by [L2]. By [A1], [L3], and [L4], $W$ has a finite basis drawn from $C - c_{0}$ . If $W = {0}$ , then $W$ and the singleton $A$ are closed directly. Otherwise its positive-dimensional coordinate map pulls the Euclidean norm back to a norm on $R^{d}$ ; [L5] and [L6] show that a convergent sequence in $W$ has its limit in $W$ . Thus [L1] and [L8] make $W$ , and hence $A$ , closed in every case. Therefore $C$ and $\overline{C}$ have the same affine hull.

A1A2F1L1L2L3L4L5L6L8givenalgebra

2.1

The basis vectors from step 1.1 give points of $C$ whose simplex has a positive barycentric core, so $C$ has nonempty relative interior. Fix a relative ball $B_{A} (p, r) \subseteq C$ and $y \in \overline{C}$ . For $q = (1 - t) p + t y$ with $0 \leq t < 1$ , [A2] and [L1] give a sequence $y_{k} \in C$ tending to $y$ ; choose a term so close that $t ∥ y - y_{k} ∥_{2} < (1 - t) r / 2$ . If $h \in W$ and $∥ h ∥_{2} < (1 - t) r / 2$ , then $p_{h} = p + \frac{t (y - y_{k}) + h}{1 - t}$ lies in $A$ and within distance $r$ of $p$ , hence lies in $B_{A} (p, r)$ , while $q + h = (1 - t) p_{h} + t y_{k} \in C$ by [F1]. Thus a relative ball about every strict segment point lies in $C$ , so every such point belongs to $ri C$ .

step 1.1A2F1L1L2L3L4choosealgebra

3.1

Let $x \in ri (\overline{C})$ . If $x \neq p$ , choose small $ε > 0$ such that $y = x + ε (x - p)$ remains in a relative ball of $\overline{C}$ about $x$ ; then $x = \frac{ε}{1 + ε} p + \frac{1}{1 + ε} y,$ so step 2.1 gives $x \in ri C$ . The case $x = p$ and the reverse inclusion are immediate. If $W = R^{n}$ , relative and ordinary interiors agree. If $W$ is proper, choose $v \notin W$ ; every ambient ball about $A$ contains a point displaced by a small nonzero multiple of $v$ and hence outside $A$ , so both ordinary interiors are empty. Thus $int C = int \overline{C}$ .

step 1.1step 2.1L4choosealgebra

4.1

By [L8], $\overline{\overline{C}} = \overline{C}$ . Combining this common closure with step 3.1 and the boundary formula [L7] gives $\partial C = \partial \overline{C}$ .

step 3.1L7L8algebra∎

Depends on

Used by

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85 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

D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 4 (standard reference, not scraped)