# Relationships between affine closures and convex closures

Let $V$ denote a vector space. Then the following concepts make sense:

• affine subset of $V$
• affine closure (affine "hull") of a subset of $V$

Suppose $V$ is in fact a real vector space. Then the following concepts also make sense:

• convex subset of $V$
• convex closure (convex "hull") of a subset of $V$.

Now given a subset $A \subseteq V$, write $\mathrm{aff}(A)$ for the affine closure and $\mathrm{con}(A)$ for the convex closure. Clearly, both $\mathrm{aff}$ and $\mathrm{con}$ are finitary closure operators. Furthermore, we have $\mathrm{aff}(A) \supseteq \mathrm{con}(A)$. This implies that $\mathrm{con}(\mathrm{aff}(A)) = \mathrm{aff}(A)$, or in other words that every affine subset is convex. However, these are very superficial relationships between $\mathrm{aff}$ and $\mathrm{con}$.

Question 0. Does anyone know of an article or book that goes beyond the observation that $\mathrm{aff}(A) \supseteq \mathrm{con}(A)$ to actually list all the important elementary relationships between these two closure operators?

Intuitively, it seems that if $A$ and $B$ are convex subsets of $V$, then if their intersection is sufficiently "large" or "non-trivial", then $\mathrm{aff}(A) = \mathrm{aff}(B)$. For example, suppose two line segments intersect at a point. Well this is quite trivial, so we cannot expect their affine closures to agree. On the other hand, suppose the intersection of two line segments is a non-trivial line segment; in particular assume their intersection has cardinality greater than or equal to $2$. Then their affine closures should agree.

Question 1. How can we formalize the intuition that if two convex sets have a sufficiently "non-trivial" intersection, then their affine closures necessarily agree?

Bonus points if you can do it without using the concepts of "basis" or "dimension."

• Just curious, why do you need $V$ to be a real vector space before you talk about convex subsets or convex closures? Both of those concepts are well defined for non-real vector spaces as well. – Michael Grant Apr 26 '15 at 23:12
• @MichaelGrant, a subset $X$ of $V$ is convex iff for all $x,y \in X$ and all $a,b \in \mathbb{R}$, we have: $(a \geq 0) \wedge (b \geq 0) \wedge (a+b=1) \rightarrow ax+by \in X$. But those order relations don't make sense over an arbitrary field. So technically, what we need is for $V$ to be a vector space over an ordered field, or even an $R$-module where $R$ is a partially-ordered commutative ring. Edit. The reason we can talk about convex subsets of $\mathbb{C}$ is because $\mathbb{C}$ can be viewed as a $2$-dimensional real vector space. – goblin Apr 27 '15 at 0:02

Question 1. This intuition is not correct in general. For example, let $A$ be a filled square that is $\{(x_1, x_2) \mid \left| x_1\right| \leq 1 , \left|x_2\right| \leq 1\}$ and $B$ be a line that intersects with $A$ (for instance, $B = \{(x_1, x_2) \mid x_1=0\}$). In this case affine hull of $A$ is $\mathbb{R}^2$, while affine hull of $B$ is $\mathbb{R}$.
• In the second case, I would regard the intersection of $A$ and $B$ as trivial relative to $A$. The intuition is surely correct for a sufficiently carefully chosen notion of "trivial." Also, I would prefer to say that the affine hull of $B$ is $\mathbb{R} \times \{0\}$, as opposed to $\mathbb{R}$. I'll check out Dimitri's book when I get a chance. – goblin Apr 30 '15 at 6:18