<p>The continuity axiom can be interpreted as follows: Let $(\textbf{x}^n)$ and $(\textbf{y}^n)$ be two sequence of goods, such that $(\textbf{x}^n)$ converges to $\textbf{x}$ and $(\textbf{y}^n)$ to $\textbf{x}$. Then the continuity axiom says that if $\forall n$ $(\textbf{x}^n)$$\succsim$$(\textbf{y}^n)$, then $\textbf{x}$$\succsim$$\textbf{y}$ also holds true(here, $(\textbf{x}^n)$,$(\textbf{y}^n),\textbf{x}, \textbf{y} \in\mathbb{R_+^n} $). This statement is equivalent to the fact that the upper contour set and lower contour set is closed (think of $(\textbf{x}^n)$ and $(\textbf{y}^n)$ to be convergent sequences of the upper and lower contour sets respectively,i.e;$(\textbf{x}^n)\in (upper contour)$, and $(\textbf{y}^n)\in (lower contour)$, $\forall n$. Then, the sequences should converge within their respective sets).</p>

<p>The above statement, however, does not hold true for lexicographic preferences. To see that, consider any two bundles of $\mathbb{R_+^2} $represented by $((1/2)^n,0) \&amp; (0,k)$ where, $k&gt;0$. It is routine to check that the bundle $((1/2)^n,0)$ converges to $(0,0)$. Now, according to the concept of lexicographic preference(I hope you're familiar with it), $((1/2)^n,0)\succsim(0,k)$. However, the preferences change as the bundle $((1/2)^n,0)$ converges to $(0,0)$(i.e; now the bundle $(0,k)\succsim(0,0)$,as n tends to an infinite value). Thus, lexicographic preferences violate the continuity axiom.</p>

CC BY-SA 4.0
2018-09-13T16:12:21.480Z
Thank you for great explanation! It couldn't have been made more clear.
Elina Gilbert
CC BY-SA 4.0
2018-09-13T15:56:20.237Z
2018-09-13T15:56:20.237Z
superhulk
<p>Take $y_1=y_2=1$. As the figure below shows, the lower left segment of the boundary of the upper contour set of the lexicographic preference is open.</p>

<p><a href="https://i.sstatic.net/2uzi1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2uzi1.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2018-09-13T15:41:31.440Z
2018-09-13T15:52:09.103Z
2018-09-13T15:52:09.103Z
Herr K.
<p>In the following picture two preference relations are defined - </p>

<p>Ist preference relation defines lexicographic relation. Though we know that lexicographic preferences doesn't have utility function. They are defined at a point so we cannot define a utility function - is discontinuous. But how can we show it in terms of upper and lower contour set being closed?</p>

<p>Why second one is continuous? </p>

<p>Note - some terminologies - </p>

<p>A binary relation >~ on the metric space X is continuous if, for all x in X, the upper and lower contour sets, { y in X : y > ~ x} and {y in X : x >~ y}, respectively, are closed. </p>

<p>The upper contour set of x is >~ (x) = {y in X : y >~ x} </p>

<p>The lower contour set of x is >~ (x) = {y in X : x>~ y} </p>

<p><img src="https://i.sstatic.net/HshMx.jpg" alt="enter image description here"></p>

<microeconomics><utility>
Continuous function in microeconomics
Elina Gilbert
# Continuous function in microeconomics

(Asked by: username_0 on 2018-09-13)

In the following picture two preference relations are defined -

Ist preference relation defines lexicographic relation. Though we know that lexicographic preferences doesn't have utility function. They are defined at a point so we cannot define a utility function - is discontinuous. But how can we show it in terms of upper and lower contour set being closed?

Why second one is continuous?

Note - some terminologies -

A binary relation >~ on the metric space X is continuous if, for all x in X, the upper and lower contour sets, { y in X : y > ~ x} and {y in X : x >~ y}, respectively, are closed.

The upper contour set of x is >~ (x) = {y in X : y >~ x}

The lower contour set of x is >~ (x) = {y in X : x>~ y}

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

**Tags:** microeconomics, utility

---

## 2 answers

### Answer by username_2 on 2018-09-13. Score: 4 | Accepted answer

Take $y\_1=y\_2=1$. As the figure below shows, the lower left segment of the boundary of the upper contour set of the lexicographic preference is open.

[![enter image description here](https://i.sstatic.net/2uzi1.png)](https://i.sstatic.net/2uzi1.png)


---

### Answer by username_1 on 2018-09-13. Score: 1

The continuity axiom can be interpreted as follows: Let $(\textbf{x}^n)$ and $(\textbf{y}^n)$ be two sequence of goods, such that $(\textbf{x}^n)$ converges to $\textbf{x}$ and $(\textbf{y}^n)$ to $\textbf{x}$. Then the continuity axiom says that if $\forall n$ $(\textbf{x}^n)$$\succsim$$(\textbf{y}^n)$, then $\textbf{x}$$\succsim$$\textbf{y}$ also holds true(here, $(\textbf{x}^n)$,$(\textbf{y}^n),\textbf{x}, \textbf{y} \in\mathbb{R\_+^n} $). This statement is equivalent to the fact that the upper contour set and lower contour set is closed (think of $(\textbf{x}^n)$ and $(\textbf{y}^n)$ to be convergent sequences of the upper and lower contour sets respectively,i.e;$(\textbf{x}^n)\in (upper contour)$, and $(\textbf{y}^n)\in (lower contour)$, $\forall n$. Then, the sequences should converge within their respective sets).

The above statement, however, does not hold true for lexicographic preferences. To see that, consider any two bundles of $\mathbb{R\_+^2} $represented by $((1/2)^n,0) \& (0,k)$ where, $k>0$. It is routine to check that the bundle $((1/2)^n,0)$ converges to $(0,0)$. Now, according to the concept of lexicographic preference(I hope you're familiar with it), $((1/2)^n,0)\succsim(0,k)$. However, the preferences change as the bundle $((1/2)^n,0)$ converges to $(0,0)$(i.e; now the bundle $(0,k)\succsim(0,0)$,as n tends to an infinite value). Thus, lexicographic preferences violate the continuity axiom.

> **Comments:**
> **username_0**: Thank you for great explanation! It couldn't have been made more clear.  

---

