<p>Single-peakedness of preferences is not enough to guarantee the existence of a <a href="https://en.wikipedia.org/wiki/Condorcet_criterion" rel="nofollow noreferrer">Condorcet winner</a>, a position that cannot be defeated in majority voting (hence preventing cycling). You would also need that the set of alternatives to admit a linear order (and single-peakedness is with respect to this order). </p>

<p>With a linear order on alternatives and a profile of single-peaked preferences, it is easy to argue that the "median voter", a position that divides the set of voters into two subsets of equal size based on whether their bliss points are on the left or right of the median position, is always a Condorcet winner. For a more rigorous presentation, I'd refer you to Proposition 21.D.1 and its proof in the <a href="http://global.oup.com/ushe/product/microeconomic-theory-9780195073409" rel="nofollow noreferrer">MWG textbook</a>. </p>

<hr>

<p>To see why the linear order on alternatives is necessary, consider the following figure (taken from <a href="https://en.wikipedia.org/wiki/Condorcet_paradox" rel="nofollow noreferrer">Wikipedia</a>):</p>

<p><a href="https://i.sstatic.net/st2iU.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/st2iU.png" alt="enter image description here"></a></p>

<p>Here, the space of alternatives is 2-dimensional, and there is no natural way to define a linear order on <span class="math-container">$\mathbb R^2$</span>. Voters' preferences are still single-peaked in the sense that each voter has only one bliss point. As a result, we cannot rule out cycling (non-existence of Condorcet winner) as is shown in the example.</p>

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2019-04-11T22:25:50.770Z
2019-04-11T22:54:57.360Z
2019-04-11T22:54:57.360Z
Herr K.
<p>How can you prove the theorem that Single peaked preferences imply that there are no Condorcet cycles</p>

<political>
Single Peaked Preferences and Condorcet Cycles
PhysicsEnthu
# Single Peaked Preferences and Condorcet Cycles

(Asked by: username_0 on 2019-04-11)

How can you prove the theorem that Single peaked preferences imply that there are no Condorcet cycles

**Tags:** political

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## 1 answers

### Answer by username_1 on 2019-04-11. Score: 3

Single-peakedness of preferences is not enough to guarantee the existence of a [Condorcet winner](https://en.wikipedia.org/wiki/Condorcet_criterion), a position that cannot be defeated in majority voting (hence preventing cycling). You would also need that the set of alternatives to admit a linear order (and single-peakedness is with respect to this order).

With a linear order on alternatives and a profile of single-peaked preferences, it is easy to argue that the "median voter", a position that divides the set of voters into two subsets of equal size based on whether their bliss points are on the left or right of the median position, is always a Condorcet winner. For a more rigorous presentation, I'd refer you to Proposition 21.D.1 and its proof in the [MWG textbook](http://global.oup.com/ushe/product/microeconomic-theory-9780195073409).

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To see why the linear order on alternatives is necessary, consider the following figure (taken from [Wikipedia](https://en.wikipedia.org/wiki/Condorcet_paradox)):

[![enter image description here](https://i.sstatic.net/st2iU.png)](https://i.sstatic.net/st2iU.png)

Here, the space of alternatives is 2-dimensional, and there is no natural way to define a linear order on $\mathbb R^2$. Voters' preferences are still single-peaked in the sense that each voter has only one bliss point. As a result, we cannot rule out cycling (non-existence of Condorcet winner) as is shown in the example.


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