<p>Yes, compactness is sufficient and no need to restrict yourself to subsets of <span class="math-container">$\mathbb{R}$</span>.</p>
<p>The proof is by contradiction. Assume that <span class="math-container">$\succeq$</span> has no greatest element. Consider the sets:
<span class="math-container">$$
V_x = \{y \in X| x \succ y\}
$$</span>
These are open sets as <span class="math-container">$\succeq$</span> is continuous. If <span class="math-container">$(V_x)_{x \in X}$</span> do not cover <span class="math-container">$X$</span> then there is an <span class="math-container">$y$</span> that is no set <span class="math-container">$V_x$</span> for all <span class="math-container">$x$</span>. This means that for all <span class="math-container">$x \in X$</span>, <span class="math-container">$x \not \succ y$</span>, so <span class="math-container">$y \succeq x$</span> for all <span class="math-container">$x$</span>, which means that <span class="math-container">$y$</span> is the greatest element a contradiction.</p>
<p>From this, it follows that <span class="math-container">$(V_x)_{x \in X}$</span> is an open cover of <span class="math-container">$X$</span>. By compactness, it has a finite subcover, say <span class="math-container">$V_{x_1},\ldots, V_{x_n}$</span>. By completeness and transitivity, we can find <span class="math-container">$x_i \in \{x_1, \ldots, x_n\}$</span> such that <span class="math-container">$x_i \succeq x_j$</span> for all <span class="math-container">$j \in \{1,\ldots, n\}$</span>.</p>
<p>Now for any <span class="math-container">$y \in X$</span> either <span class="math-container">$y = x_j$</span> for some <span class="math-container">$j$</span> which means <span class="math-container">$x_i \succeq y = x_j$</span> or <span class="math-container">$y \in V_{x_j}$</span> for some <span class="math-container">$j$</span> which means that:
<span class="math-container">$$
x_i \succeq x_j \succ y.
$$</span>
So <span class="math-container">$x_i$</span> is a greatest element, which gives the desired contradiction.</p>

CC BY-SA 4.0
2021-09-20T15:45:12.330Z
2021-09-20T15:45:12.330Z
tdm
<p>I have a compact subset of <span class="math-container">$\mathbb{R}$</span>, <span class="math-container">$X$</span>. An agent has a continuous, transitive and complete preference relation <span class="math-container">$\succsim$</span> over <span class="math-container">$X$</span>. I am wondering whether there exists a <span class="math-container">$y\in X$</span> such that <span class="math-container">$y\succsim x$</span> for all <span class="math-container">$x\in X$</span>. I have the following so far:</p>
<p>If <span class="math-container">$X$</span> were simply closed, then the answer would be no. This is because we could define the preference relation: <span class="math-container">$x\succsim y$</span> <em>iff</em> <span class="math-container">$x\geq y$</span> on <span class="math-container">$\mathbb{R}$</span> (which is a closed set). Clearly, there is no maximal element.</p>
<p>But I am not sure about the case in which <span class="math-container">$X$</span> is compact.</p>
<p>Thank you</p>

CC BY-SA 4.0
2021-09-20T15:36:59.627Z
Hint: Use [extreme value theorem](https://en.wikipedia.org/wiki/Extreme_value_theorem#Generalization_to_metric_and_topological_spaces) and the result that a continuous, transitive and complete preference relation can be represented by a continuous utility function.
Herr K.
<microeconomics><mathematical-economics><consumer-theory>
Existence of maximal element in a compact choice set
Paul
# Existence of maximal element in a compact choice set

(Asked by: username_0 on 2021-09-20)

I have a compact subset of $\mathbb{R}$, $X$. An agent has a continuous, transitive and complete preference relation $\succsim$ over $X$. I am wondering whether there exists a $y\in X$ such that $y\succsim x$ for all $x\in X$. I have the following so far:

If $X$ were simply closed, then the answer would be no. This is because we could define the preference relation: $x\succsim y$ *iff* $x\geq y$ on $\mathbb{R}$ (which is a closed set). Clearly, there is no maximal element.

But I am not sure about the case in which $X$ is compact.

Thank you

**Tags:** microeconomics, mathematical-economics, consumer-theory
> **Comments:**
> **username_1**: Hint: Use [extreme value theorem](https://en.wikipedia.org/wiki/Extreme_value_theorem#Generalization_to_metric_and_topological_spaces) and the result that a continuous, transitive and complete preference relation can be represented by a continuous utility function.  

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## 1 answers

### Answer by username_2 on 2021-09-20. Score: 2

Yes, compactness is sufficient and no need to restrict yourself to subsets of $\mathbb{R}$.

The proof is by contradiction. Assume that $\succeq$ has no greatest element. Consider the sets:
$$
V\_x = \{y \in X| x \succ y\}
$$
These are open sets as $\succeq$ is continuous. If $(V\_x)\_{x \in X}$ do not cover $X$ then there is an $y$ that is no set $V\_x$ for all $x$. This means that for all $x \in X$, $x \not \succ y$, so $y \succeq x$ for all $x$, which means that $y$ is the greatest element a contradiction.

From this, it follows that $(V\_x)\_{x \in X}$ is an open cover of $X$. By compactness, it has a finite subcover, say $V\_{x\_1},\ldots, V\_{x\_n}$. By completeness and transitivity, we can find $x\_i \in \{x\_1, \ldots, x\_n\}$ such that $x\_i \succeq x\_j$ for all $j \in \{1,\ldots, n\}$.

Now for any $y \in X$ either $y = x\_j$ for some $j$ which means $x\_i \succeq y = x\_j$ or $y \in V\_{x\_j}$ for some $j$ which means that:
$$
x\_i \succeq x\_j \succ y.
$$
So $x\_i$ is a greatest element, which gives the desired contradiction.


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