# infimum - notational problem

I have divided a two-dimensional coordinate system into several regions that form a partition. Now I want to define a line that is formed by the lower bound of some of these regions, and I cannot think of a correct, let alone elegant way to formulate this.

For example, I have three regions $A,B,C$, where $A=\left\{(x,y)|x<y-2,x<1\right\}$, and $B=\left\{(x,y)|x<y-2,x\geq1\right\}$, and $C=\mathbb{R}²\setminus (A\cup B)$, i.e. $A$ is the region above the line $l(x)=y=x+2$ and to the left of $x=1$, and $B$ is the region above the line $l$ and to the right of $X=1$, and $C$ is the region below it. Now in my case, I have no possibility of writing just $l(x)=y=x+2$, because (it's a macroeconomical model) I'm not dealing with numbers, but with symbols on the axes. I need to define the line as being the lower bound of region $A\cup B$.

My idea is something like $$l=\left\{ \left( x,\inf_y\vert(x,y)\in A \cup B\right) \right\}$$ So in words: the line is the set of all points where y is the smallest element s.th. $(x,y)$ is still contained n $A\cup B$.

Is this even correct, and is this the most elegant solution? Is there any difference to $$l= \inf_y \left\{ \left( x,y\right) \right\}\vert (x,y)\in A\cup B$$ Do I maybe have to work with $\min$ instead of $\inf$? I'm wondering because the notation $\inf_y(x,y)$ suggests that $(x,y)$ is the smallest element, but clearly a 2-dimensional vector is not in an ordered set.

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It is rather $\inf\{y\mid (x,y)\in A\cup B\}$ for a given $x$.
so in total, the equation will be $l=\left\{ (x,\inf\left\{y\vert \exists x: (x,y) \in A \cup B\right\}) \right\}$? – Marie. P. Oct 6 '12 at 11:23
ah no, it will be $l=\left\{ (x,\inf\left\{y\vert (x,y) \in A \cup B\right\}) \right\}$. now it's clear. – Marie. P. Oct 6 '12 at 11:31
yes. $\min$ is justified only if you know that $\min$ exists, else rather stay at $\inf$. – Berci Oct 6 '12 at 11:34