Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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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.

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

Invertible group is open and inversion is continuous

Statement

Let A be a unital complex Banach algebra (Unital Banach algebra) and let a∈A× be invertible with inverse a−1 (Invertible element and general linear group of a Banach algebra). Then:

  1. every b∈A with ∥a−1∥ ∥b−a∥<1 is invertible, with b−1=(1−a−1(a−b))−1a−1and∥b−1∥≤∥a−1∥1−∥a−1∥ ∥b−a∥;
  2. A× is an open subset of A;
  3. inversion A×→A×, a↦a−1, is continuous at every point of A× (with the relative topology on A×).

Facts & Assumptions

Given: A unital complex Banach algebra A, an invertible a∈A×, and an element b∈A with ∥a−1∥ ∥b−a∥<1. Put x:=a−1(a−b), so that b=a(1−x).

[L1]

The norm on A is submultiplicative, ∥1∥=1, multiplication is associative and bilinear, and the norm is continuous with respect to itself: ∥uv∥≤∥u∥ ∥v∥ and ∥u−u′∥≤∥u−u′′∥+∥u′′−u′∥ (Unital Banach algebra).

[L2]

An element c∈A is invertible exactly when it has a two-sided inverse, which is then unique; inverses satisfy (uv)−1=v−1u−1 for invertible u,v, and (u−1)−1=u (Invertible element and general linear group of a Banach algebra).

[L3]

If ∥y∥<1 then 1−y is invertible with (1−y)−1=∑n≥0yn and every tail bound ∥(1−y)−1−∑n≤Nyn∥≤∥y∥N+1/(1−∥y∥); in particular ∥(1−y)−1∥≤1/(1−∥y∥) (Neumann series).

Proof

technique · direct
1.1

The element x:=a−1(a−b) satisfies ∥x∥≤∥a−1∥ ∥a−b∥=∥a−1∥ ∥b−a∥<1 by [L1], and b=a−(a−b)=a−aa−1(a−b)=a(1−x), where the middle step uses aa−1=1 from [L2].

L1L2algebra
1.2

For the difference of inverses one has the algebraic identity b−1−a−1=b−1(a−b)a−1 whenever both inverses exist, because b−1(a−b)a−1=b−1aa−1−b−1ba−1=b−1−a−1, using [L2].

L2L1algebra
2.1

By [L3] applied to x with ∥x∥<1, the element 1−x is invertible with (1−x)−1=∑n≥0xn and ∥(1−x)−1∥≤1/(1−∥x∥)≤1/(1−∥a−1∥ ∥b−a∥).

step 1.1L3
3.1

Since b=a(1−x) with both factors invertible, [L2] gives that b is invertible with b−1=(1−x)−1a−1=(1−a−1(a−b))−1a−1, and taking norms with [L1] and [step 2.1] gives ∥b−1∥≤∥(1−x)−1∥ ∥a−1∥≤∥a−1∥/(1−∥a−1∥ ∥b−a∥); this is claim 1.

step 2.1L2L1algebra
4.1

Claim 2 follows: given a∈A×, every b with ∥b−a∥<1/∥a−1∥ satisfies the hypothesis verified in [step 3.1] and hence lies in A×, so A× contains the open ball of that radius about a.

step 3.1L1
4.2

In particular, whenever ∥b−a∥<1/(2∥a−1∥) the bound of [step 3.1] gives ∥b−1∥≤∥a−1∥/(1−12)=2∥a−1∥.

step 3.1algebra
5.1

Combining [step 1.2] with [step 4.2] and [L1], for ∥b−a∥<1/(2∥a−1∥) one has ∥b−1−a−1∥≤∥b−1∥ ∥b−a∥ ∥a−1∥≤2∥a−1∥2∥b−a∥, which tends to 0 as b→a; this is claim 3.

step 1.2step 4.2L1algebra
6.1

Claims 1, 2 and 3 are exactly the three assertions of the statement, so the theorem is proved.

step 3.1step 4.1step 5.1∎

Depends on

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Sources