I have been able to show this constructively, but would like to prove it by induction. However, I am stuck with the induction step:

Consider $\succsim$ defined over $X=\{x_1,...,x_n\}$ and where without loss of generality $x_1$ is the element such that $x_1 \succsim x, \forall x \in X$. Now, consider adding element $x_{n+1}$ to the set. My reasoning is that since $\succsim$ is defined over the original $X$, there is nothing telling us that $x_{n+1}$ should be in a complete relation with the elements in $X$ (i.e. $\succsim$ is not defined over a bigger set).

Is this reasoning correct? In that case, I don't see how you could prove this by induction.


1 Answer 1


You are almost there, but I think you are starting in the wrong place. I think the key is as follows: we want to prove the claim for set $X=\{x_1,\ldots,x_n\}$.

Start with some $X'\subseteq{X}$ for which the claim is true. Such an $X'$ exists because the claim is true for the set $\{x_1\}$ (by completeness and reflexivity).

If $X'=X$ then we're done. If $X'\neq X$ then we proceed with induction:

Denote by $\bar{x}$ an $x\in X'$ such that $\bar{x}\succsim y, \forall\ y\in X'$.

Now construct a new set $X''=X'\cup\{x_i\}$, where $x_i\in X\backslash X'$. Because $\succsim$ is rational (preferences are complete) over $X$, we either have $x_i\succsim \bar{x}$ or $\bar{x}\succsim x_i$ or both. Thus, if the claim holds for $X'$ it also holds for $X''$.

We can repeat this inductive step until $X''=X$, which must happen after finitely many iterations because $X$ is finite.

  • $\begingroup$ Thank you very much. Induction on subsets makes sense. $\endgroup$
    – econ86
    Oct 1, 2019 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.