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

  • 1
    $\begingroup$ Hint: Use extreme value theorem and the result that a continuous, transitive and complete preference relation can be represented by a continuous utility function. $\endgroup$
    – Herr K.
    Sep 20, 2021 at 15:36

1 Answer 1


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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