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For an explicit basis extension in ${\mathbb{R}}^2$, start with $A=\{u\}$, where $u=(1,1)$. Its linear span is the line $y=x$. Adding $w=(1,-1)$ preserves linear independence, because $w$ is outside that line, and gives a basis. Every vector has the finite linear combinationThe vector $2u=(2,2)$ cannot be added while preserving linear independence: it is already in the linear span. Many choices outside the line work, so maximality does not select a unique basis.

For a finite-dimensional vector space with a supplied finite spanning set, a greedy algorithm scans the list, retaining each vector precisely when it is outside the linear span of those already retained. Starting with a linearly independent $A$, the same scan adjoins only the needed vectors. Retained vectors remain linearly independent, and every skipped vector is in the final linear span, so the result is a basis containing $A$. Finite reasoning needs no axiom of choice.

For an arbitrary vector space, this finite stopping argument is unavailable. Accepting Zorn's lemma asserts existence of a maximal element once every chain in a partial order has an upper bound; it supplies no general computational recipe. Over ZF, Zorn's lemma, the axiom of choice and the well-ordering theorem are equivalent. Particular infinite vector spaces can still have explicit bases, such as $1,x,x^2,\dots$ for the vector space of univariate polynomials over a specified field; infinity alone does not force a nonconstructive argument.^[ref]

Suppose for contradiction that the vector space of univariate polynomials $P$ were complete in the given norm. For $n\ge 0$, letEach $P_n$ is finite-dimensional and therefore closed by the fact that a finite-dimensional subspace is closed. It is a proper linear subspace of $P$, so it has empty interior: if a linear subspace contained any open ball, translation and scalar multiplication would make it contain every vector. Hence every $P_n$ is a nowhere dense set.

But every polynomial has finite degree, soThis writes the supposed complete metric space $P$ as a countable union of nowhere-dense closed subsets, contradicting the Baire category theorem. Thus no norm can make $P$ complete. Equivalently, this is the countable-basis obstruction that a Banach space has uncountable Hamel dimension.