# Continuity and Leximin

Consider a relation $\geq$ over the set of real-valued vectors. We say that $\geq$ is continuous if for any positive integer $n$, and any $\pi\in\mathbb Z^n,u\in\mathbb R ^n$ we have that the sets $\{v\in\mathbb R ^n:(\pi,v)\geq(\pi,u)\}$ and $\{v\in\mathbb R ^n:(\pi,v)\leq(\pi,u)\}$ are closed, where $(\pi,v)$ indicates the vector formed by appending $v$ to $\pi$.

(This is kind of a weird definition, but I think it's implied by [but isn't identical to?] this one.)

The Leximin ordering is one where $x\geq y$ if the smallest value of $x$ is greater than the smallest value of $y$. If the smallest values are equal, we compare the second smallest, etc.

Blackorby, Bossert and Donaldson claim that the leximin ordering is not continuous. I'm struggling to understand why.

For example, consider $S=\{v:v\geq (0,0)\}$. This is the set $\{x:x_i\geq 0\}$. That set seems closed to me. For example, there doesn't seem to be anything less than $(0,0)$ itself.

What am I not understanding?

## 1 Answer

Given $\pi \in \mathbb{Z}^n$ and $u \in \mathbb{R}^n$, by $U[\pi,u]$ I will denote the set $$\{ v \in \mathbb{R}^n : ( \pi , u ) \leq ( \pi , v ) \}.$$

Fixing $\pi = \langle 0,0, \ldots , 0 \rangle \in \mathbb{Z}^n$, and let $u = \langle 1 , 3 , \ldots , 3 \rangle \in \mathbb{R}^n$, note that for each $j > 0$ the $n$-tuple $v_j = \langle 1 + \frac{1}{j} , 2, 2, \ldots , 2 \rangle$ belongs to the set $U[\pi,u]$. Also note that the $v_j$ converge to $v = \langle 1 ,2,2, \ldots , 2 \rangle$, but $v \notin U_\leq[\pi,u]$. It follows that $U[\pi,u]$ is not closed. (Note that I have implicitly assumed that $n > 1$; for $n=1$ the sets will be closed.)

(If you disallow tuples with equal entries, this example can be easily modified so that each tuple which appears has distinct entries.)