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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."

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  • $\begingroup$ 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. $\endgroup$ – Michael Grant Apr 26 '15 at 23:12
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    $\begingroup$ @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. $\endgroup$ – goblin GONE Apr 27 '15 at 0:02
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Question 0. I do not know any article in this context. However, there are very good problems at the end of the first chapter of Dimitri Bertsikas Convex Analysis and Optimization book that goes over many useful properties of affine hulls and convex hulls.

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 $\mathbb{R}$.

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  • $\begingroup$ 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. $\endgroup$ – goblin GONE Apr 30 '15 at 6:18

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