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In reading about convex optimization, the author states that all convex sets are affine. Are affinity and convexity equivalent? If I understand, both definitions incorporate the notion that a set is affine/convex iff for every two points in the set, the line connecting them is also in the set. It seems as if they are as their respective definitions are almost identical. I would appreciate a delineation of the differences between affine sets and convex sets.

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[I am not at all an expert, but] I think the noun form of affine is affineness, not affinity. –  Srivatsan Dec 6 '11 at 2:48
    
@Srivatsan : I got away with "affinity" in a paper published in the Monthly in 2003. –  Michael Hardy Dec 6 '11 at 2:58
    
@Michael, In fact, I might be totally wrong :-); I cannot find any authoritative sources for "affineness". –  Srivatsan Dec 6 '11 at 3:00
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@Srivatsan: In a mathematical context affineness is preferable. In a genealogical context only affinity is correct. –  Brian M. Scott Dec 6 '11 at 3:09
    
That one paper that I wrote is the only "authoritative" source I have for "affinity", and I used it only because it was moderately amusing and not demonstrably wrong. –  Michael Hardy Dec 7 '11 at 20:35
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3 Answers

up vote 7 down vote accepted

A set $S$ is convex iff for every pair of points $x,y\in S$, the line segment $\overline{xy}$ joining $x$ to $y$ is a subset of $S$. $S$ is affine iff for every pair of points $x,y\in S$, the whole infinite line containing $x$ and $y$ is a subset of $A$. In the xy-plane, for instance, $S=\{(x,0):0\le x\le 1\}$ is a convex set but not an affine set: the smallest affine set containing $S$ is the whole $x$-axis.

Mathematically, $S$ is affine iff it contains every affine combination of its points, where an affine combination of the points $x_1,\dots,x_n\in S$ is any point of the form $$a_1x_1+a_2x_2+\cdots+a_nx_n$$ such that $$a_1+a_2+\cdots+a_n=1\;.$$

$S$ is convex iff it contains every convex combination of its points. Convex combinations are the special case of affine combinations in which all of the coefficients are non-negative. That is, a convex combination of the points $x_1,\dots,x_n\in S$ is any point of the form

$$a_1x_1+a_2x_2+\cdots+a_nx_n$$ such that $$a_1+a_2+\cdots+a_n=1$$ and $$a_1,a_2,\dots,a_n\ge 0\;.$$

You can think of a convex combination of points as a kind of weighted average of those points; an affine combination is then a weighted ‘average’ in which some of the weights are allowed to be negative. In particular, $$\{ax+by:a+b=1\text{ and }a,b\ge 0\}$$ is simply the set of points on the line segment from $x$ to $y$; the point $ax+yb$ is $b$ fraction of the way from $x$ to $y$. The set

$$\{ax+by:a+b=1\}\;,$$

on the other hand, includes the whole line through $x$ and $y$. The point $-x+2y$, for instance, does not lie between $x$ and $y$: instead, $y$ is between $x$ and $-x+2y$.

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Convexity condition is somewhat subset of condition for being affine.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Tomas Aug 9 '13 at 15:38
    
It would be better to elaborate your answer. –  TZakrevskiy Aug 9 '13 at 15:54
    
It sounds as if you're saying that convex $\implies$ affine, but since you are talking about conditions for being such, I think you actually mean the opposite. It would be nice if you were to expand on this answer a bit. –  robjohn Aug 9 '13 at 17:09
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You have got everything wrong. Check it here. http://homepages.rpi.edu/~mitchj/handouts/affine/.

In fact, every affine set is convex and not vice versa.

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