<p>It would be with strict preference, otherwise it could always be proven to not hold, by taking <span class="math-container">$a=b$</span>, or even any two points where <span class="math-container">$a \sim b$</span>.</p>
<p>Well, then that would be true simply because it’s the definition of continuous preference.</p>
<p>An example would be literally any continuous utility function over its domain, such as <span class="math-container">$x+y$</span>, <span class="math-container">$x^\frac{1}{3} y^\frac{2}{3}$</span>, <span class="math-container">$\log(x) + \log(y), \text{min}\{x,y\}$</span> and many more.</p>
<p>It is much more interesting to think about non-examples.</p>
<p>One non-example is <span class="math-container">$U(x,y) = x + ⌊y⌋$</span> where <span class="math-container">$⌊•⌋$</span> is the floor function, which rounds down the argument to the previous integer.</p>
<p>Let’s work on the Euclidean metric <span class="math-container">$d(•,•)$</span> over <span class="math-container">${\mathbb{R}^+}^2$</span>.</p>
<p>Let’s take <span class="math-container">$a = (1,1)$</span> and <span class="math-container">$b = (1,0)$</span>.</p>
<p>We have <span class="math-container">$U(a) = 2 &gt; 1 = U(b)$</span>, therefore <span class="math-container">$a \succ b$</span></p>
<p>Now let <span class="math-container">$\epsilon \in (0,2) $</span> be arbitrary.</p>
<p>Let <span class="math-container">$x = (1-\frac{\epsilon}{2},1-\frac{\epsilon}{2})$</span>.</p>
<p>We have that <span class="math-container">$d(x,a) = \sqrt{2 (\frac{\epsilon}{2})^2} = \frac{\epsilon}{\sqrt{2}} &lt; \epsilon$</span>, hence <span class="math-container">$x \in N_\epsilon (a)$</span>.</p>
<p>Note <span class="math-container">$1-\frac{\epsilon}{2} \in (0,1)$</span>, hence <span class="math-container">$⌊1-\frac{\epsilon}{2}⌋ = 0$</span>.</p>
<p>Therefore, <span class="math-container">$U(x) = 1-\frac{\epsilon}{2} &lt; 1 = U(b)$</span>, hence <span class="math-container">$b \succ x$</span>.</p>
<p>Note a point always belongs to its own neighborhood, so we just found <span class="math-container">$(a,b) \in {\mathbb{R}^+}^2 $</span> with <span class="math-container">$a \succ b$</span>, such that for arbitrarily small <span class="math-container">$\epsilon, \delta &gt; 0$</span> we can always find <span class="math-container">$x$</span> (and <span class="math-container">$y = b$</span>) such that <span class="math-container">$x \in N_\epsilon (a)$</span> and <span class="math-container">$y \in N_\delta (b)$</span>, but <span class="math-container">$y \succ x$</span>.</p>
<p>Therefore, we can conclude that the preferences represented by this utility function are not continuous.</p>
<p>The same can be proven for lexicographic preferences for any points of the form <span class="math-container">$a = (c,a_2), b = (c,b_2)$</span>, where <span class="math-container">$a_2 &gt; b_2$</span>, as <span class="math-container">$(c-\frac{\epsilon}{2},a_2) \in N_\epsilon (a)$</span> for arbitrarily small <span class="math-container">$\epsilon &gt;0$</span>, as <span class="math-container">$a \succ b$</span> but <span class="math-container">$b \succ (c-\frac{\epsilon}{2},a_2)$</span>.</p>
<p>Some purposes of continuous preferences are to be able to draw nice indifference curves and for a Walrasian Equilibrium to exist in pure exchange economies. The latter also requiring strict convexity and strict monotonicity as well.</p>

CC BY-SA 4.0
2023-08-27T19:11:51.183Z
2023-08-29T10:16:35.447Z
2023-08-29T10:16:35.447Z
Nicolas Torres
<p>&quot;A preference is continuous if for any <span class="math-container">$a,b\in X$</span> with <span class="math-container">$a\succsim b$</span> there are some neighborhoods <span class="math-container">$N_{\varepsilon}(a)$</span>, <span class="math-container">$N_{\delta}(b)$</span> around <span class="math-container">$a$</span> and <span class="math-container">$b$</span> such that for every <span class="math-container">$x \in N_{\varepsilon}(a)$</span>, <span class="math-container">$y\in N_{\delta}(b)$</span> we have <span class="math-container">$x\succsim y$</span>&quot;</p>
<p>Is that true and what is the purpose of continuous preference? Please give me an example.</p>

<microeconomics><mathematical-economics><preferences><definition>
Continuity of preference
Huy Lê Thanh
# Continuity of preference

(Asked by: username_0 on 2023-08-26)

"A preference is continuous if for any $a,b\in X$ with $a\succsim b$ there are some neighborhoods $N\_{\varepsilon}(a)$, $N\_{\delta}(b)$ around $a$ and $b$ such that for every $x \in N\_{\varepsilon}(a)$, $y\in N\_{\delta}(b)$ we have $x\succsim y$"

Is that true and what is the purpose of continuous preference? Please give me an example.

**Tags:** microeconomics, mathematical-economics, preferences, definition

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## 1 answers

### Answer by username_1 on 2023-08-27. Score: 3

It would be with strict preference, otherwise it could always be proven to not hold, by taking $a=b$, or even any two points where $a \sim b$.

Well, then that would be true simply because it’s the definition of continuous preference.

An example would be literally any continuous utility function over its domain, such as $x+y$, $x^\frac{1}{3} y^\frac{2}{3}$, $\log(x) + \log(y), \text{min}\{x,y\}$ and many more.

It is much more interesting to think about non-examples.

One non-example is $U(x,y) = x + ⌊y⌋$ where $⌊•⌋$ is the floor function, which rounds down the argument to the previous integer.

Let’s work on the Euclidean metric $d(•,•)$ over ${\mathbb{R}^+}^2$.

Let’s take $a = (1,1)$ and $b = (1,0)$.

We have $U(a) = 2 > 1 = U(b)$, therefore $a \succ b$

Now let $\epsilon \in (0,2) $ be arbitrary.

Let $x = (1-\frac{\epsilon}{2},1-\frac{\epsilon}{2})$.

We have that $d(x,a) = \sqrt{2 (\frac{\epsilon}{2})^2} = \frac{\epsilon}{\sqrt{2}} < \epsilon$, hence $x \in N\_\epsilon (a)$.

Note $1-\frac{\epsilon}{2} \in (0,1)$, hence $⌊1-\frac{\epsilon}{2}⌋ = 0$.

Therefore, $U(x) = 1-\frac{\epsilon}{2} < 1 = U(b)$, hence $b \succ x$.

Note a point always belongs to its own neighborhood, so we just found $(a,b) \in {\mathbb{R}^+}^2 $ with $a \succ b$, such that for arbitrarily small $\epsilon, \delta > 0$ we can always find $x$ (and $y = b$) such that $x \in N\_\epsilon (a)$ and $y \in N\_\delta (b)$, but $y \succ x$.

Therefore, we can conclude that the preferences represented by this utility function are not continuous.

The same can be proven for lexicographic preferences for any points of the form $a = (c,a\_2), b = (c,b\_2)$, where $a\_2 > b\_2$, as $(c-\frac{\epsilon}{2},a\_2) \in N\_\epsilon (a)$ for arbitrarily small $\epsilon >0$, as $a \succ b$ but $b \succ (c-\frac{\epsilon}{2},a\_2)$.

Some purposes of continuous preferences are to be able to draw nice indifference curves and for a Walrasian Equilibrium to exist in pure exchange economies. The latter also requiring strict convexity and strict monotonicity as well.


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