<p>I think I just found a counterexample to your teacher's claim.</p>

<p>Suppose completeness <em>is violated completely</em> (sorry for the pun): $L\not\succsim L'$ and $L'\not\succsim L$ for any $L,L'\in\mathcal L$, i.e. no two lotteries are comparable. </p>

<p>Then I claim that transitivity and continuity are both satisfied vacuously. </p>

<p><strong>Transitivity</strong>: $L\succsim L'$ and $L'\succsim L''$ imply $L\succsim L''$ for any $L,L',L''\in\mathcal L$. But since no two lotteries are comparable by $\succsim$, the antecedent is always false, and so transitivity is always satisfied. </p>

<p><strong>Continuity</strong>: for any $L,L',L''\in\mathcal L$, the sets $\{\alpha\in[0,1]:\alpha L+(1-\alpha)L'\succsim L''\}$ and $\{\alpha\in[0,1]:L''\succsim\alpha L+(1-\alpha)L'\}$ are both closed. Again, since no two lotteries are comparable, both sets are empty and therefore closed. Hence, continuity is satisfied in a vacuous sense.</p>

CC BY-SA 3.0
2017-03-04T21:09:56.037Z
@denesp: Thanks for pointing it out. It's corrected.
Herr K.
CC BY-SA 3.0
2017-03-04T20:39:00.153Z
2017-03-04T21:09:13.180Z
2017-03-04T21:09:13.180Z
Herr K.
<p>My teacher mentioned off-hand that the first Von Neumann Morgenstern axiom of completeness is not actually necessary to prove the Expected Utility Property, and that the axiom of completeness is in fact implied by the axioms of continuity and transitivity. I don't see why this is. </p>

<expected-utility>
How can the VNM completeness axiom be derived from the transitivity and continuity axioms?
gcrunnels
# How can the VNM completeness axiom be derived from the transitivity and continuity axioms?

(Asked by: username_0 on 2017-03-04)

My teacher mentioned off-hand that the first Von Neumann Morgenstern axiom of completeness is not actually necessary to prove the Expected Utility Property, and that the axiom of completeness is in fact implied by the axioms of continuity and transitivity. I don't see why this is.

**Tags:** expected-utility

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## 1 answers

### Answer by username_1 on 2017-03-04. Score: 2

I think I just found a counterexample to your teacher's claim.

Suppose completeness *is violated completely* (sorry for the pun): $L\not\succsim L'$ and $L'\not\succsim L$ for any $L,L'\in\mathcal L$, i.e. no two lotteries are comparable.

Then I claim that transitivity and continuity are both satisfied vacuously.

**Transitivity**: $L\succsim L'$ and $L'\succsim L''$ imply $L\succsim L''$ for any $L,L',L''\in\mathcal L$. But since no two lotteries are comparable by $\succsim$, the antecedent is always false, and so transitivity is always satisfied.

**Continuity**: for any $L,L',L''\in\mathcal L$, the sets $\{\alpha\in[0,1]:\alpha L+(1-\alpha)L'\succsim L''\}$ and $\{\alpha\in[0,1]:L''\succsim\alpha L+(1-\alpha)L'\}$ are both closed. Again, since no two lotteries are comparable, both sets are empty and therefore closed. Hence, continuity is satisfied in a vacuous sense.

> **Comments:**
> **username_1**: @denesp: Thanks for pointing it out. It's corrected.  

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