<p><strong>Definition</strong>: <span class="math-container">$\succsim$</span> on <span class="math-container">$\mathbb{R}^2_+$</span> is said to be continuous if</p>
<p>for any two convergent sequences <span class="math-container">$(x'_n,y'_n), (x''_n,y''_n) \in \mathbb{R}^2_+$</span>,</p>
<p><span class="math-container">$(x'_n,y'_n)\succsim (x''_n,y''_n)$</span> for all <span class="math-container">$n\in\mathbb{N}$</span></p>
<p>implies</p>
<p><span class="math-container">$\lim_{n\rightarrow\infty} (x'_n,y'_n)\succsim \lim_{n\rightarrow\infty} (x''_n,y''_n)$</span></p>
<p><strong>Definition:</strong> Lexicographic preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$\mathbb{R}^2_+$</span> is defined as follows:</p>
<p>for any <span class="math-container">$(x_1,y_1), (x_2,y_2) \in \mathbb{R}^2_+$</span>,</p>
<p><span class="math-container">$(x_1,y_1)\succsim (x_2,y_2)$</span>
<strong>if and only if</strong>
either <span class="math-container">$x_1 &gt; x_2$</span> or <span class="math-container">$(x_1 = x_2 \text{ and } y_1\geq y_2)$</span></p>
<p><strong>Claim.</strong> Lexicographic preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$\mathbb{R}^2_+$</span> is not continuous.</p>
<p><strong>Proof.</strong> Consider the following two sequences:</p>
<p><span class="math-container">$(x'_n,y'_n) = \left(1+\frac{1}{n},0\right)$</span></p>
<p>and a constant sequence</p>
<p><span class="math-container">$(x''_n,y''_n) = (1,1)$</span></p>
<p>Observe that <span class="math-container">$(x'_n,y'_n)\succsim (x''_n,y''_n)$</span> for all <span class="math-container">$n\in\mathbb{N}$</span> holds, but <span class="math-container">$\lim_{n\rightarrow\infty} (x'_n,y'_n) = (1,0) \not\succsim (1,1)=\lim_{n\rightarrow\infty} (x''_n,y''_n)$</span>.</p>
<p><a href="https://i.sstatic.net/D0rhx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/D0rhx.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-09-28T06:00:22.363Z
2023-09-28T07:42:25.900Z
2023-09-28T07:42:25.900Z
Amit
<p>For example for the lexicographix preferences on <span class="math-container">$[0,1] \times [0,1]$</span>take any ball around <span class="math-container">$(1,1)$</span>. Although you know that <span class="math-container">$(1,1) \succ (1,0)$</span>, there exist points in the given ball (<span class="math-container">$(1-\epsilon,1)$</span>) such that <span class="math-container">$(1,0)\succ (1-\epsilon,1)$</span> which contradicts with the definition of continuity. I hope this helps!</p>

CC BY-SA 4.0
2023-09-27T23:51:20.163Z
2023-09-27T23:51:20.163Z
Dennis Papadakis
<p>I’m having trouble understanding why lexicographic preference relations aren’t continuous. I need inspiration on how this proof would work.</p>

CC BY-SA 4.0
2023-09-15T17:56:40.113Z
https://economics.stackexchange.com/search?q=lexicographic
Herr K.
<microeconomics><preferences>
Lexicographic preference relation not continuous
GraceLynn87
# Lexicographic preference relation not continuous

(Asked by: username_0 on 2023-09-15)

I’m having trouble understanding why lexicographic preference relations aren’t continuous. I need inspiration on how this proof would work.

**Tags:** microeconomics, preferences

> **Comments:**
> **username_1**: https://economics.stackexchange.com/search?q=lexicographic  

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## 2 answers

### Answer by username_2 on 2023-09-28. Score: 3

**Definition**: $\succsim$ on $\mathbb{R}^2\_+$ is said to be continuous if

for any two convergent sequences $(x'\_n,y'\_n), (x''\_n,y''\_n) \in \mathbb{R}^2\_+$,

$(x'\_n,y'\_n)\succsim (x''\_n,y''\_n)$ for all $n\in\mathbb{N}$

implies

$\lim\_{n\rightarrow\infty} (x'\_n,y'\_n)\succsim \lim\_{n\rightarrow\infty} (x''\_n,y''\_n)$

**Definition:** Lexicographic preference relation $\succsim$ on $\mathbb{R}^2\_+$ is defined as follows:

for any $(x\_1,y\_1), (x\_2,y\_2) \in \mathbb{R}^2\_+$,

$(x\_1,y\_1)\succsim (x\_2,y\_2)$
**if and only if**
either $x\_1 > x\_2$ or $(x\_1 = x\_2 \text{ and } y\_1\geq y\_2)$

**Claim.** Lexicographic preference relation $\succsim$ on $\mathbb{R}^2\_+$ is not continuous.

**Proof.** Consider the following two sequences:

$(x'\_n,y'\_n) = \left(1+\frac{1}{n},0\right)$

and a constant sequence

$(x''\_n,y''\_n) = (1,1)$

Observe that $(x'\_n,y'\_n)\succsim (x''\_n,y''\_n)$ for all $n\in\mathbb{N}$ holds, but $\lim\_{n\rightarrow\infty} (x'\_n,y'\_n) = (1,0) \not\succsim (1,1)=\lim\_{n\rightarrow\infty} (x''\_n,y''\_n)$.

[![enter image description here](https://i.sstatic.net/D0rhx.png)](https://i.sstatic.net/D0rhx.png)


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### Answer by username_3 on 2023-09-27. Score: 1

For example for the lexicographix preferences on $[0,1] \times [0,1]$take any ball around $(1,1)$. Although you know that $(1,1) \succ (1,0)$, there exist points in the given ball ($(1-\epsilon,1)$) such that $(1,0)\succ (1-\epsilon,1)$ which contradicts with the definition of continuity. I hope this helps!


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