# Easy convex sets question

Question 2.23 out of Boyd and Vanderberghe:

Give an example of two closed convex sets that are disjoint but cannot be strictly separated.

The obvious idea is to take something like unbounded sets which are disjoint but approach each other in the limit. For example, f(x) = 1/x and g(x) = -1/x. But isn't x=0 a strictly separating hyperplane here?

Thanks!

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To say that two convex sets $C$ and $D$ in $\mathbb{R}^n$ can be separated by a hyperplane $\mathcal{H}$ is to say that there is a linear functional $f$ on $\mathbb{R}^n$ such that $\mbox{sup} \{f\left(x\right)\: : \: x\in C\} \leq \mbox{inf} \{f\left(x\right)\: : \: x \in D\}$. $\mathcal{H}$ will then be the hyperplane $\{x \in \mathbb{R}^n \: : \: f\left(x\right) = c\}$, which is a coset of the kernel of $f$.

We say that $C$ and $D$ can be strictly separated if a linear functional $f$ can be chosen so that the above inequality is strict. Your intuition about what sorts of closed convex sets can be separated but not strictly separated is correct. This can happen when $C$ and $D$ are not compact and approach arbitrarily close to each other without meeting. So if you let $C$ be the points in $\mathbb{R}^2$ such that $y \geq \frac{1}{x}$ with $x > 0$ and $D$ be the points such that $y \leq -\frac{1}{x}$ with $x > 0$, then $C$ and $D$ are separated by the functional that projects onto the $y$ axis, but they are not strictly separated by this functional.

Noah's answer gives another good example. joriki's answer clears up a lot of things, and the paper he links to is a good introduction to this topic.

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Take $X = \{(x,y) \mid xy\geq 1, x,y>0\}$ and $Y = \{(x,y) \mid x\leq 0\}$.

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• You both seem to have ignored the "convex" part of the question. What you need is not the graphs of these functions but the sets they bound. Also you need to select one of the two branches of each of $1/x$ and $-1/x$ in order to get a convex set.