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Let's say that A is a convex set in $R^2$. Now assume that L is a line in $R^2$.

$L=\{x: p\cdot x = t\}$ where p and x are both contained in $R^2$, $p\cdot x$ is the inner product of p and x, and t is any number in R.

Also A and L do NOT intersect. It seems obvious to me that A is within an open halfspace of L, but I'm not sure what a proof might look like. I guess I'd have to show that all points in A will be greater than or less than an equivalent point in L, but I haven't a clue how to go about it. Any help? Thanks.

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up vote 3 down vote accepted

Assume that A contains one point each in both open halfspaces of $L$. The segment connecting those two points intersects $L$, and since $A$ is a convex set, it contains that point, too, in contradiction to the fact that $A$ and $L$ don't intersect. Hence $A$ contains points of at most one open halfspace of $L$.

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Well that was simple. Thanks :) – user9554 Apr 13 '11 at 21:05

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