TITLE: Disjoint convex sets that are not strictly separated
QUESTION [19 upvotes]: Question 2.23 out of Boyd & Vanderberghe's Convex Optimization:

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, $f(x) = \frac1x$ and $g(x) = -\frac1x$. But isn't $x=0$ a strictly separating hyperplane here?

REPLY [5 votes]: I'll just add that, for a slightly different definition of convex sets being "strictly separated", the example given by Albert Seppi doesn't work. (My definition is that $C$ and $D$ convex sets in $\mathbb{R}^n$ are strictly separated if there is a hyperplane $a^Tx=\beta$ such that $a^Ty<\beta$ for all $y\in C$ and $a^Ty>\beta$ for all $y\in D$)
In the latter case, $x=0$ separates strictly $\{(x,y)\mid y\geq \frac1x, x>0\}$ and $\{(x,y)\mid y\leq -\frac1x, x>0\}$.
However, it is impossible to separate the sets $\{(x,y)\mid y\geq \frac1x, x>0\}$ and $\{(x,y)\mid y\leq0\}$ because any line of slope non-zero will intersect the second set, so such a line must be of the form $(1\mbox{ } 0)^Tx=\beta$ for some $\beta$. But if $\beta\leq0$, it intersects the second set whereas $\beta >0$ implies that the line will intersect the first set.