Bases and Dimension: Examples and Counterexamples
About this page
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.
✓ 8 results · all verified · 7 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.
1 · Prerequisites
Abstract Algebra
Linear Algebra
Real Analysis
2 · Logical flowchart
3 · Items
Example
The vector (1,2) ∈ ℝ² has coordinate list (1,2) in the standard ordered basis, (2,1) in its reversal, and (2,-1) in the ordered basis ((1,1),(1,0))
Example
The standard unit families eₖ ∈ F^ℕ form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle
Example
ℝ as a vector space over ℚ has a basis, and every such basis is infinite; the existence proof exhibits none
Counterexample
The standard unit families { eᵢ : i ∈ ℕ } are linearly independent in F^ℕ but do not span it: the constant family 1_F is not a finite linear combination of them
Counterexample
{(1,0), (0,1), (1,1)} spans F² and is linearly dependent, so a spanning set need not be a basis; each of its three two-element subsets is a basis
Counterexample
Inside the space of eventually zero families, the linear subspace spanned by { eᵢ : i ≥ 1 } is proper and has a basis equinumerous with a basis of the whole space, so "equal dimension forces equality" fails without finite dimension
Counterexample
Three distinct lines U₀, U₁, U₂ in F² have dim_F(U₀+U₁+U₂) = 2 while the inclusion-exclusion analogue of the dimension formula predicts 3, so the two-subspace formula does not extend
False statement
FALSE: the union of two linearly independent subsets of a vector space is linearly independent
Sources
- Basis (linear algebra) (Wikipedia)
- Coordinate vector (Wikipedia)
- Sequence space (Wikipedia)
- Cambridge University Press excerpt: Vector spaces and bases
- Hamel basis (Wikipedia)
- Axiom of choice (Wikipedia)
- University of Vermont notes: Infinite-dimensional vector spaces
- Linear independence (Wikipedia)
- Auburn University notes: Spanning sets and linear independence
- Dimension (vector space) (Wikipedia)
- Dimension theorem for vector spaces (Wikipedia)
- Inclusion-exclusion principle (Wikipedia)
- UC Berkeley Math 110 notes: Linear algebra
- Jim Hefferon, Linear Algebra: Answers to exercises