On learning about infinite dimensional vector spaces, one learns that we need to use the axiom of choice to assert that every such vector space has a basis; indeed, it's equivalent to the AoC to assert this. However, I had not known any "natural" examples of such a vector space till I studied the proof of the barvinok algorithm. I produce the example here.

Consider a space such as $S \equiv \mathbb R^3$. Now, consider the vector space spanned by the indicator functions of polyhedra in $S$. that's a mouthful, so let's break it down.

A polyhedra is defined as a set of points that is defined by linear inequalities: $P \equiv \{ x \in S : a_i \cdot x \leq b_i, i \in [1\dots n] \}$, for all $a_i \in S$, $b \in \mathbb R$.

The indicator functions are of the form:

we can define a vector space of these functions over $\mathbb R$, using the "scaling" action as the action of $\mathbb R$ on these functions:

The vector space $V$ is defined as the span of the indicator functions of all polyhedra. It's clearly a vector space, and a hopefully intuitive one. However, note that the set we generated this from (indicators of polyhedra) don't form a basis since they have many linear dependencies between them. For example, one can write the equation:

*---*   *-*   *-*   *|###|   |#|   |#|   ||###| = |#| + |#| - ||###|   |#|   |#|   |*---*   *-*   *-*   *