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 $\mathbb{N}$

Let $V$ be a vector space over a field $F$ (Vector space over a field) and suppose $V$ has a spanning subset $S$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$) with $S \approx n$ for some $n \in \mathbb{N}$ (Equinumerous sets, $A \approx B$ and $A \preceq B$). Then:

1. every linearly independent subset $L \subseteq V$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent) is finite (Finite, countably infinite, countable, uncountable), and the unique $m \in \mathbb{N}$ with $L \approx m$ satisfies $m \le n$;
2. no linearly independent subset of $V$ is equinumerous with $\mathbb{N}$.

• Claim 2 is the form in which later items say a space is infinite-dimensional. Exhibiting a linearly independent subset equinumerous with $\mathbb{N}$ shows, by this corollary read backwards, that the space has no finite spanning set at all, hence no finite basis. That is exactly the route taken on the companion page by the explicit infinite basis for the eventually zero families and by the independent set of $F^{\mathbb{N}}$ that does not span it.

• The bound is on the independent set, not on the spanning set. A spanning set may be enlarged freely without ceasing to span, so no bound in the other direction holds; what is bounded is how many vectors can be independent, and the bound is the size of any finite spanning set.

• Nothing here assumes $V$ has a basis. The hypothesis is a finite spanning set, which need not be independent; that a spanning set contains a basis is Every spanning subset of a vector space contains a basis, proved later and by a different route.