The standard unit families $\{\, e_i : i \in \mathbb{N} \,\}$ are linearly independent in $F^{\mathbb{N}}$ but do not span it: the constant family $1_F$ is not a finite linear combination of them

False claim: if $V$ is an infinite-dimensional vector space over a field $F$ (Finite-dimensional vector space, and its dimension $\dim_F V$; infinite-dimensional means having no finite basis) and $B \subseteq V$ is a linearly independent subset that is not finite, then $\operatorname{span}(B) = V$.

Take $F$ any field, $V := F^{\mathbb{N}}$ the function space of all families $x : \mathbb{N} \to F$ with the pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$), and $B := \{\, e_i : i \in \mathbb{N} \,\}$ the standard unit families, where $e_i(i) = 1_F$ and $e_i(n) = 0_F$ for $n \ne i$. Then $B$ is linearly independent and not finite, $V$ is infinite-dimensional, and yet $\operatorname{span}(B) = E$, the linear subspace of eventually zero families, which is not all of $V$: the constant family $c$ with $c(n) = 1_F$ for every $n$ lies outside it.

So $B$ is an infinite linearly independent set that is not a basis of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), although it is a basis of $E$.

• What is true instead. $B$ extends to a basis of $F^{\mathbb{N}}$ by Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $\operatorname{span}(S) = V$, there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$, applied with $L = B$ and $S = F^{\mathbb{N}}$. That is not in tension with anything above: the extension adds vectors outside $\operatorname{span}(B)$, and it is produced by Zorn's lemma rather than exhibited.

• Where the finite-dimensional intuition breaks. In a space of dimension $n$, an independent set with $n$ elements does span, because If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$ forbids enlarging it and For $B \subseteq V$ the following are equivalent: $B$ is a basis; $B$ is a maximal linearly independent subset of $V$; $B$ is a minimal spanning subset of $V$ — maximality and minimality being in the inclusion order then makes it a basis. Infinitude is not a substitute for maximality: $B$ is infinite and still extendable.

• The whole content of the failure is that a linear combination is finite. Every element of $\operatorname{span}(B)$ is built from finitely many of the $e_i$ ($\operatorname{span}(S)$ is exactly the set of linear combinations of finite lists of elements of $S$, and $\operatorname{span}(\varnothing) = \{0_V\}$), hence vanishes from some index on, while $c$ vanishes nowhere.