TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$

Statement

Let $n \in N$ with $n \geq 1$ , let $(x_{k})$ be a sequence in $R^{n}$ whose series converges (Series of vectors in $R^{n}$ , absolute convergence, rearrangement, and the set of rearrangement sums) and write $s : = \sum_{k = 0}^{\infty} x_{k}$ . Let $Γ$ and $Γ^{⊥}$ be as in The subspace $Γ$ of directions along which a series converges absolutely, and its orthogonal complement $Γ^{⊥}$ . Then:

Nonemptiness. $s \in S (x)$ , so $S (x) \neq \emptyset$ .
Containment. $S (x) \subseteq s + Γ^{⊥},$ the affine subspace through $s$ with direction $Γ^{⊥}$ (The subspace $Γ$ of directions along which a series converges absolutely, and its orthogonal complement $Γ^{⊥}$ ). Equivalently, $t - s \in Γ^{⊥}$ for every rearrangement sum $t$ .
The absolutely convergent case. If $\sum x_{k}$ converges absolutely then $Γ = R^{n}$ , $Γ^{⊥} = {0}$ , the affine subspace is the single point ${s}$ , and $S (x) = {s}$ .
The one-dimensional conditionally convergent case. Let $n = 1$ and identify $R^{1}$ with $R$ as in Series of vectors in $R^{n}$ , absolute convergence, rearrangement, and the set of rearrangement sums. If $\sum x_{k}$ converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index) then $Γ = {0}$ , $Γ^{⊥} = R^{1}$ , and the containment of clause 2 is an equality, $S (x) = s + Γ^{⊥} = R^{1}$ , by the published The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ .

What this theorem does not say, stated here and repeated in the Remarks. It proves a containment and nothing more. Whether $S (x)$ is all of $s + Γ^{⊥}$ when $n \geq 2$ is not settled anywhere on this page, and no item on this page asserts anything about it in either direction. Clause 4 is the case $n = 1$ , where the answer is supplied by a published theorem about the real line; it is not evidence for any statement in higher dimensions.

Facts & Assumptions

Given: A natural $n \geq 1$ ; a sequence $(x_{k})$ in $R^{n}$ with $\sum x_{k}$ convergent of sum $s$ ; a bijection $σ$ of $N$ ; a vector $a \in Γ$ ; the partial sums $s_{N} = \sum_{k < N} x_{k}$ and $s_{N}^{σ} = \sum_{k < N} x_{σ (k)}$ .

[L1]

Series of vectors, absolute convergence, rearrangement, $S (x)$ , and the identification of $R^{1}$ with $R$ (Series of vectors in $R^{n}$ , absolute convergence, rearrangement, and the set of rearrangement sums, Rearrangement of a series along a bijection of $N$ , and unconditional convergence, Injection, surjection, bijection, Isometry, isometric embedding, and the subspace metric on a subset).

[L2]

$Γ$ and $Γ^{⊥}$ are linear subspaces; $a \in Γ$ means $\sum_{k} ∣ ⟨ a, x_{k} ⟩ ∣$ converges; $Γ = R^{n}$ exactly when $\sum x_{k}$ converges absolutely; and $s + W$ denotes the coset of a linear subspace $W$ (The subspace $Γ$ of directions along which a series converges absolutely, and its orthogonal complement $Γ^{⊥}$ , Linear subspace of a vector space).

[L3]

The inner product is bilinear and symmetric, $⟨ y, y ⟩ = ∥ y ∥_{2}^{2}$ , positive definiteness gives $⟨ y, y ⟩ = 0$ only for $y = 0$ , and Cauchy-Schwarz gives $∣ ⟨ a, y ⟩ ∣ \leq ∥ a ∥_{2} ∥ y ∥_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L4]

Laws of finite sums and the induction principle (Laws of finite sums and finite products, Finite sums and finite products, by recursion, The principle of mathematical induction); finite sums in $R^{n}$ are pointwise (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1).

[L5]

Dirichlet's rearrangement theorem: an absolutely convergent real series has, for every bijection $σ$ of $N$ , a rearrangement converging to the same sum (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum, Absolutely convergent and conditionally convergent series, and the general starting index).

[L6]

The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement converging to $c$ (The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ clause 1); and over $R$ a convergent series is absolutely convergent or conditionally convergent and not both (For a series of real numbers, unconditional convergence and absolute convergence are the same property, Absolutely convergent and conditionally convergent series, and the general starting index).

[L7]

Convergence in $(R^{n}, d_{2})$ and in $R$ , uniqueness of limits, and the componentwise criterion (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Limits and Cauchy sequences of reals, A sequence in a metric space has at most one limit, For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm).

[L8]

An absolutely convergent series in $R^{n}$ converges, every rearrangement converges to the same sum, and $S (x)$ is then a single point (An absolutely convergent series in $R^{n}$ converges, and every rearrangement converges to the same sum).

[L9]

Absolute value and order arithmetic: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ \geq 0$ , and $u > 0$ gives $u^{- 1} > 0$ (Basic properties of the absolute value, Inverses of positives are positive, and reciprocation reverses order).

Proof

technique · direct

1.1L1

The identity map of $N$ is a bijection and the rearrangement along it is the original series, so $s \in S (x)$ and clause 1 holds.

1.2L3L4

For every $a \in R^{n}$ and every finite list $u : p \to R^{n}$ , $⟨ a, \sum_{j < p} u_{j} ⟩ = \sum_{j < p} ⟨ a, u_{j} ⟩$ : at $p = 0$ both sides are $0$ , and the successor step is additivity of the inner product in its second argument.

1.3L3L7L9

If $u_{N} \to u$ in $(R^{n}, d_{2})$ then $⟨ a, u_{N} ⟩ \to ⟨ a, u ⟩$ in $R$ , since $∣ ⟨ a, u_{N} ⟩ - ⟨ a, u ⟩ ∣ = ∣ ⟨ a, u_{N} - u ⟩ ∣ \leq ∥ a ∥_{2} ∥ u_{N} - u ∥_{2}$ , so a tolerance $ε / (∥ a ∥_{2} + 1)$ on the right serves for $ε$ on the left.

1.4L1L2L9

Now let $n = 1$ and suppose $\sum x_{k}$ converges conditionally, so the real series $\sum_{k} (x_{k})_{0}$ converges and $\sum_{k} ∣ (x_{k})_{0} ∣$ diverges. For $a \in R^{1}$ , $⟨ a, x_{k} ⟩ = a_{0} (x_{k})_{0}$ and $∣ ⟨ a, x_{k} ⟩ ∣ = ∣ a_{0} ∣ ∣ (x_{k})_{0} ∣$ ; if $a_{0} \neq 0$ then convergence of $\sum_{k} ∣ a_{0} ∣ ∣ (x_{k})_{0} ∣$ would give convergence of $\sum_{k} ∣ (x_{k})_{0} ∣$ after multiplying by the positive $1 / ∣ a_{0} ∣$ , which is false, so $a \in Γ$ forces $a_{0} = 0$ ; and $a = 0$ does lie in $Γ$ . Hence $Γ = {0}$ .

2.1step 1.2step 1.3L1L7

Let $t \in S (x)$ , say $s_{N}^{σ} \to t$ for a bijection $σ$ , and let $a \in Γ$ . By steps 1.2 and 1.3, $⟨ a, t ⟩ = \lim_{N} ⟨ a, s_{N}^{σ} ⟩ = \lim_{N} \sum_{k < N} ⟨ a, x_{σ (k)} ⟩$ , so the real series $\sum_{k} ⟨ a, x_{σ (k)} ⟩$ converges with sum $⟨ a, t ⟩$ .

2.2step 1.2step 1.3L7

In the same way $⟨ a, s ⟩ = \lim_{N} ⟨ a, s_{N} ⟩ = \lim_{N} \sum_{k < N} ⟨ a, x_{k} ⟩$ , so $\sum_{k} ⟨ a, x_{k} ⟩$ converges with sum $⟨ a, s ⟩$ .

2.3step 1.4L2L3

With $Γ = {0}$ the condition defining $Γ^{⊥}$ is $⟨ 0, y ⟩ = 0$ , which holds for every $y$ , so $Γ^{⊥} = R^{1}$ and $s + Γ^{⊥} = R^{1}$ .

3.1step 2.1step 2.2L2L5

The real sequence $k \mapsto ⟨ a, x_{σ (k)} ⟩$ is the rearrangement along $σ$ of the sequence $k \mapsto ⟨ a, x_{k} ⟩$ , and the latter series converges absolutely because $a \in Γ$ ; so by Dirichlet's theorem the two series have the same sum.

3.2step 1.4step 2.3L1L6L7

By the Riemann series theorem applied to the conditionally convergent real series $\sum_{k} (x_{k})_{0}$ , every real $c$ is the sum of some rearrangement of it; transporting along the identification of $R$ with $R^{1}$ , every element of $R^{1}$ lies in $S (x)$ . So $S (x) = R^{1} = s + Γ^{⊥}$ , which with steps 1.4 and 2.3 is clause 4.

4.1step 2.1step 2.2step 3.1L3

Combining steps 2.1, 2.2 and 3.1 gives $⟨ a, t ⟩ = ⟨ a, s ⟩$ , hence $⟨ a, t - s ⟩ = 0$ by bilinearity.

5.1step 4.1L2

Since $a \in Γ$ was arbitrary, $t - s \in Γ^{⊥}$ , that is $t \in s + Γ^{⊥}$ ; as $t \in S (x)$ was arbitrary, clause 2 holds.

6.1step 5.1L2L3L8

Suppose $\sum x_{k}$ converges absolutely. Then $Γ = R^{n}$ , so any $y \in Γ^{⊥}$ satisfies $⟨ y, y ⟩ = 0$ and hence $y = 0$ ; thus $Γ^{⊥} = {0}$ and $s + Γ^{⊥} = {s}$ . Moreover $S (x) = {s}$ by [L8], so clause 3 holds and the containment of clause 2 is an equality in this case.

7.1step 1.1step 5.1step 6.1step 3.2∎

Clauses 1, 2, 3 and 4 are steps 1.1, 5.1, 6.1 and 3.2.

Remarks

This theorem proves containment only, and the reverse inclusion is not proved, assumed, or asserted anywhere on this page. For $n \geq 2$ the question whether every point of $s + Γ^{⊥}$ is a rearrangement sum remains open within this page; The same question in $R^{d}$ : what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order records that proof boundary. Finite-dimensional orthogonal decomposition and convex-separation results are now published elsewhere in this library, but this page supplies no reverse-inclusion rearrangement construction or convergence estimates. See Conventions of this page, the standing $n \geq 1$ hypothesis, and what is taken up elsewhere in the reading order.
The title claims exactly clause 2 and clause 1, and no more. A title asserting that $S (x)$ is the affine subspace would assert the reverse inclusion, which is not proved here.
Clause 4 is the published one-dimensional dichotomy seen from this page. Over $R$ a convergent series is either absolutely convergent, and then $Γ$ is everything and $S$ is a point (clause 3), or conditionally convergent, and then $Γ$ is ${0}$ and $S$ is the whole line (clause 4). Both extremes are consistent with clause 2, and both are equalities; that is a fact about dimension $1$ , where a linear subspace of $R^{1}$ is ${0}$ or everything and there is no room in between.
What the containment already rules out. Even without the reverse inclusion, clause 2 forbids a rearrangement sum from leaving the affine subspace. That is enough to refute the naive $R^{n}$ analogue of the Riemann series theorem, and the companion page does so with an elementary witness, using clause 2 and nothing further.

The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The set of rearrangement sums of a convergent series in RnRn is a nonempty subset of the affine subspace s+Γ⊥s+Γ⊥

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$