<p>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 <span class="math-container">$X=\{x_1,\ldots,x_n\}$</span>. </p>

<p>Start with some <span class="math-container">$X'\subseteq{X}$</span> for which the claim is true. Such an <span class="math-container">$X'$</span> exists because the claim is true for the set <span class="math-container">$\{x_1\}$</span> (by completeness and reflexivity). </p>

<p>If <span class="math-container">$X'=X$</span> then we're done. If <span class="math-container">$X'\neq X$</span> then we proceed with induction:</p>

<p>Denote by <span class="math-container">$\bar{x}$</span> an <span class="math-container">$x\in X'$</span> such that <span class="math-container">$\bar{x}\succsim y, \forall\ y\in X'$</span>.</p>

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

<p>We can repeat this inductive step until <span class="math-container">$X''=X$</span>, which must happen after finitely many iterations because <span class="math-container">$X$</span> is finite.</p>

CC BY-SA 4.0
2019-10-01T15:25:09.210Z
Thank you very much. Induction on subsets makes sense.
econ86
CC BY-SA 4.0
2019-10-01T15:21:36.457Z
2019-10-01T15:21:36.457Z
Ubiquitous
<p>I have been able to show this constructively, but would like to prove it by induction. However, I am stuck with the induction step:</p>

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

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

<microeconomics><preferences>
Given a rational $\succsim$ over a finite set $X$, show that there exists $x \in X$ such that $x \succsim y, \forall y \in X$
econ86
# Given a rational $\succsim$ over a finite set $X$, show that there exists $x \in X$ such that $x \succsim y, \forall y \in X$

(Asked by: username_0 on 2019-10-01)

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.

**Tags:** microeconomics, preferences

---

## 1 answers

### Answer by username_1 on 2019-10-01. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: Thank you very much. Induction on subsets makes sense.  

---

