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Every vector space $V$ over a field $F$ has a basis, assuming the axiom of choice. More generally, any linearly independent subset $A\subseteq V$ extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing $A$:It is a set, being a subset of the power set of $V$, and is nonempty because $A\in {\mathcal{P}}_A$.

A short proof outline supplies both essential steps:

Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: $\{v\}$ and $\{2v\}$ for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

Order the linearly independent subsets of a vector space by inclusion. A chain's union is independent, since every finite relation is contained in one chain member. Zorn's lemma gives a maximal independent subset. A vector outside its span could be adjoined while preserving independence, so it spans. Sending finite-support coefficient families to their linear combinations gives an isomorphism from the free module on that subset to $V$. This supplies an arbitrary, possibly infinite-dimensional basis using the axiom of choice; no pre-existing basis theorem is invoked.

A vector subspace is a nonempty subset of a vector space closed under vector addition and scalar multiplication, and is therefore itself a vector space. Equivalently, whenever $u,v$ belong to the subset, so does $au+bv$ for all allowed scalars $a,b$. In particular it contains the zero vector.

For example, the plane $W=\{(x,y,z):x+y+z=0\}\subseteq {\mathbb{R}}^3$ is a vector subspace, with basis $(1,-1,0),(1,0,-1)$. Indeed every element is $(-y)(1,-1,0)+(-z)(1,0,-1)$. The parallel plane $x+y+z=1$ misses the zero vector and is an affine subspace instead. Homogeneous constraints produce vector subspaces: $W$ is the kernel of the linear map $(x,y,z)\mapsto x+y+z$. This lets us describe all solutions of a homogeneous linear equation through a basis of its kernel.

A weight basis of a Lie algebra representation is a basis consisting of weight vectors for a chosen Cartan subalgebra. When the Cartan action is diagonalizable, choose a basis separately in every weight space and combine them. This makes weight multiplicities directly visible and gives diagonal matrices for Cartan elements.

Let $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, and let $f_i$ be the dual basis. The equivariant contraction sends $t_{ij}\otimes f_k$ to ${\delta }_{ik}{e}_j+{\delta }_{jk}{e}_i$. Its kernel has a weight basis consisting of $t_{ii}\otimes f_k$ for $i\ne k$, $t_{ij}\otimes f_k$ for three distinct indices with $i<j$, and $t_{ij}\otimes f_j-\frac{1}{2}{t}_{ii}\otimes f_i$ for $i\ne j$. The three families have sizes six, three and six. This is the irreducible fifteen-dimensional highest-weight representation with highest weight $2L_1-L_3$.

The algebra over $k[t]$ defined by $yx-xy=t$ has ordered basis $x^i{y}^j$, by its differential Ore-extension construction. Its $t$-adic completion is a formal associative formal associative deformation of $k[x,y]$. In characteristic zero its first-order multiplication term is $f(a,b)={\partial }_{y}a{\partial }_{x}b$. The cocycle is not a coboundary: a degree-two coboundary on a commutative algebra is symmetric, whereas $f(y,x)-f(x,y)=1$.

For $M$ complete energy eigenstates of a one-particle Quantum Hamiltonian for identical noninteracting fermions, order their energies ${ϵ}_1\le \cdots \le {ϵ}_M$. At fixed particle number $0\le N\le M$, the Quantum Hamiltonian and the particle-number constraint areThe ground state occupies the $N$ lowest complete one-particle states. A short minimization argument suffices: moving an occupant from a higher energy to an empty lower energy decreases the total energy. This filling describes the ground state at temperature $T=0$; exclusion by itself also applies to excited quantum states and at positive temperature. The underlying thermal description is the Fermi-Dirac distribution.^[ref]

If the last occupied energy belongs to a group of $g$ one-particle states at the same energy and $q$ of them must be occupied, with all lower states filled, the ground state subspace has dimension $\left(\genfrac{}{}{0ex}{}{g}{q}\right)$. Exclusion does not select a particular basis or a particular quantum superposition in that subspace. The factor of two for an Electron spatial orbital assumes both spin projections are included; a model restricted to one spin projection has only one state per spatial orbital. A magnetic field can split the degenerate energy levels, and interactions can change the lowest-energy electron configuration. Thus this finite noninteracting filling rule is not a complete prescription for interacting chemical atoms.