<p>The question is whether rationality is implied by continuity and monotonicity. To show that this is not the case, a counterexample would suffice. We are therefore looking for an intransitive, incomplete, monotone, continuous preference relation.</p>

<p>Suppose $X=\{x\geq 0,y\geq 0:x+y=1\}$. Thus, we form preferences over points of a line from $(0,1)$ to $(1,0)$.
Consider the preference relation defined by $(1,0) \succ (.5,.5) \succ (0,1) \succ (1,0)$ which is incomplete otherwise.</p>

<h2>Rationality</h2>

<p>Rationality consist of completeness and transitivity of the preference relation, defined as follows:</p>

<h2>Completeness</h2>

<blockquote>
  <p>A preference relation is complete, if for all $x,y \in X$, we have $x\succsim y$, $y\succsim x$, or both.</p>
</blockquote>

<p>$(.5,.5)\not \succsim (.5,.5)$, thus the preference relation is not complete.</p>

<h2>Transitivity</h2>

<blockquote>
  <p>A preference relation is transitive, if $x\succsim y$ and $y \succsim z$ imply $x\succsim z$.</p>
</blockquote>

<p>$(1,0)\succsim (.5,.5)$ and $(.5,.5) \succsim (0,1)$ hold but $(1,0)\not \succsim (0,1)$, thus the preference relation is not transitive.</p>

<h2>Continuity</h2>

<blockquote>
  <p>A preference relation is continuous if for all sequences ${(x_i,y_i)}_{i=1}^{\infty}$ converging to $(x,y)$ with $\forall i: x_i \succsim y_i$ we have $x \succsim y$.</p>
</blockquote>

<p>The preference relation does not violate continuity. Consider a sequence $x_i \succsim y_i$ which converges to $x,y$. These sequences can only be such that $x_i=x$ and $y_i=y$, and $x\neq y$, since all other $x_i,y_i$ either do not converge to $x,y$, or do not fulfill $x_i \succsim y_i$. But clearly if $x_i\succsim y_i$ then $x\succsim y$.</p>

<h2>Monotonicity</h2>

<blockquote>
  <p>A preference relation is monotone, if $x \geq y$ implies $x \succsim y$.</p>
</blockquote>

<p>The $\geq$ relation considers all elements of $X$ incomparable, thus the preference relation is monotone.</p>

<p>Thus, we have an intransitive, incomplete, monotone, continuous preference relation.</p>

CC BY-SA 3.0
2015-09-21T10:49:34.990Z
I assume that $x \leq 1, y \leq 1$, but even so the definition of your relation seems incomplete. What is preferred, (0.1,0.9) or (0,1)? (And what about the other pairings?) By the $\succeq$ between (0.5,0.5) and (0,1) do you mean $\sim$?
Giskard
CC BY-SA 3.0
2015-09-24T03:35:48.583Z
Thank you for pointing out the typing error. Regarding the remaining comments about providing an incomplete relation: This is exactly the point. We are looking for a preference relation which is intransitive + incomplete, but at the same time monotone and continuous. If we start out with a preference relation that is complete, this would defeat the purpose.
HRSE
CC BY-SA 3.0
2015-09-24T06:28:21.460Z
I see. So you mean the relation is only defined exactly where you defined it. This is not always the case. E.g.: 3 < 5, but the relation also where I have not defined it.
Giskard
CC BY-SA 3.0
2015-09-25T01:34:26.407Z
a relation is always only "defined where one defines it". Formally, a relation is a subset of the cartesian product of a set. For the definition of the relation, the specification of that subset is sufficient. Thus, you can define the relation < on the real numbers such that 3<5. This will not correspond to the usual definition, but it is nonetheless a valid specification of an (incomplete) relation.
HRSE
CC BY-SA 3.0
2015-09-25T04:04:04.430Z
Alright I'll rephrase my comment: I have tought you merely gave some  examples about how your relation would work and not an exact definition but now I understand what you meant.
Giskard
CC BY-SA 3.0
2015-09-25T06:19:46.543Z
ok, i hope now it is clear
HRSE
CC BY-SA 3.0
2015-09-21T08:47:43.967Z
2015-09-25T06:19:30.283Z
2015-09-25T06:19:30.283Z
HRSE
<p>Transitivity of preferences appeals to some "intuitive" notion of "consistency of human mind" and it can be argued that any exceptions are the "exceptions to the <em>rule</em>", and so we <em>do</em> have an adequate abstract rule.  </p>

<p>In comparison, Completeness is much more of a "leap of faith". It hangs in the air, stemming from nothing, related to nothing (<strong>so the answer to your question is no</strong>). Perhaps it can be supported by some vulgar remark that "if you <em>press</em> a person enough, then he will eventually order any pair you will put in front of him, even if only to get rid of you", but of course this, while looking good in practice, will never work in theory.  </p>

<p>So we just define Completeness to exist... why? In order to avoid rather <em>unmanageable</em> issues down the road. How useful will it be to work with non-complete preferences? How useful would it be to say "I have this model, it may result, it may not, depending on whether preferences are complete or not"... what's the use of it? We would then be forced to come up with an <em>alternative</em> decision rule: "Assuming that preferences are not complete, then if the person encounters a pair that he cannot order..." -he does <em>what</em>? Flip a coin? But this would make "incompleteness" equivalent to indifference...  </p>

<p>What else? This line of thought may be very stimulating, but it is also very challenging, and certainly path-breaking, if indeed, such a path exists or can be created. (In my opinion, various theoretical explorations of the "fuzzy" variety try to find a "middle way" for exactly this problem -where they consider a situation where the person neither has complete preferences, nor is completely "frozen" when a "difficult" pair comes up).</p>

CC BY-SA 3.0
2015-09-20T20:25:02.033Z
2015-09-20T20:25:02.033Z
Alecos Papadopoulos
<p>Consider a preference relation in $\mathbb{R}^2$ such that $x=(x_1,x_2)\succsim (y_1,y_2)=y$ $\iff$ $x_1\geq y_1$ and $x_2\geq y_2$. </p>

<p>1) You might like to argue whether this preference relation is strictly monotonic and continuous. </p>

<p>2) Is the relation defined above complete?</p>

<p>Then, as a side dish, you might also reconsider your claim that continuity is the cause of transitivity.</p>

<p>Note: I just wrote this particular one with the purpose of providing a thought experiment. More in a way to challenge your understanding. I am not sure whether this example provides an answer to your question or not.</p>

CC BY-SA 3.0
2015-09-20T19:54:44.017Z
2015-09-20T20:08:36.220Z
2015-09-20T20:08:36.220Z
ramazan
<p>Let $\succsim$ be a strictly monotonic and continuous preference relation, and let $X=\mathbb{R}^{n}$ be the consumption set.</p>

<p>Is rationality of $\succsim$ implied by these conditions? </p>

<p>I think transitivity is implied by continuity. However, completeness is troubling, as there are elements $x,y \in X$ that cannot be ordered with respect to $\leq$ or $\geq$, and so we cannot use monotonicity to show that $\succsim$ is complete.</p>

<p>I have thought of constructing a sequence $x_{n}$ with $x_{1}=x$ such that $x_{n} \to y$ and either $x_{n}\succsim x_{n+1}$ or $x_{n+1} \succsim x_{n}$. Then by transitivity and continuity we could show that $x$ and $y$ can be ordered with respect to $\succsim$, but I do not think it is possible to construct such a sequence. </p>

<p>Any help would be appreciated, but please give hints and not full solutions.</p>

CC BY-SA 3.0
2015-09-20T16:21:27.533Z
Unfortunately transitivity of a relation does not follow from only continuity. Let R be the relation 'has difference strictly less than one'. On real numbers R is continuous but not transitive.
Giskard
CC BY-SA 3.0
2015-09-20T19:16:53.240Z
I'm fairly certain that monotonic and continuous preferences are not necessarily rational.
BB King
<microeconomics><consumer-theory>
Are monotonic and continuous preferences necessarily rational?
möbius
# Are monotonic and continuous preferences necessarily rational?

(Asked by: username_0 on 2015-09-20)

Let $\succsim$ be a strictly monotonic and continuous preference relation, and let $X=\mathbb{R}^{n}$ be the consumption set.

Is rationality of $\succsim$ implied by these conditions?

I think transitivity is implied by continuity. However, completeness is troubling, as there are elements $x,y \in X$ that cannot be ordered with respect to $\leq$ or $\geq$, and so we cannot use monotonicity to show that $\succsim$ is complete.

I have thought of constructing a sequence $x\_{n}$ with $x\_{1}=x$ such that $x\_{n} \to y$ and either $x\_{n}\succsim x\_{n+1}$ or $x\_{n+1} \succsim x\_{n}$. Then by transitivity and continuity we could show that $x$ and $y$ can be ordered with respect to $\succsim$, but I do not think it is possible to construct such a sequence.

Any help would be appreciated, but please give hints and not full solutions.

**Tags:** microeconomics, consumer-theory
> **Comments:**
> **username_1**: Unfortunately transitivity of a relation does not follow from only continuity. Let R be the relation 'has difference strictly less than one'. On real numbers R is continuous but not transitive.  
> **username_2**: I'm fairly certain that monotonic and continuous preferences are not necessarily rational.  

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## 3 answers

### Answer by username_5 on 2015-09-20. Score: 8 | Accepted answer

Consider a preference relation in $\mathbb{R}^2$ such that $x=(x\_1,x\_2)\succsim (y\_1,y\_2)=y$ $\iff$ $x\_1\geq y\_1$ and $x\_2\geq y\_2$.

1) You might like to argue whether this preference relation is strictly monotonic and continuous.

2) Is the relation defined above complete?

Then, as a side dish, you might also reconsider your claim that continuity is the cause of transitivity.

Note: I just wrote this particular one with the purpose of providing a thought experiment. More in a way to challenge your understanding. I am not sure whether this example provides an answer to your question or not.


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### Answer by username_3 on 2015-09-21. Score: 4

The question is whether rationality is implied by continuity and monotonicity. To show that this is not the case, a counterexample would suffice. We are therefore looking for an intransitive, incomplete, monotone, continuous preference relation.

Suppose $X=\{x\geq 0,y\geq 0:x+y=1\}$. Thus, we form preferences over points of a line from $(0,1)$ to $(1,0)$.
Consider the preference relation defined by $(1,0) \succ (.5,.5) \succ (0,1) \succ (1,0)$ which is incomplete otherwise.

## Rationality

Rationality consist of completeness and transitivity of the preference relation, defined as follows:

## Completeness

> A preference relation is complete, if for all $x,y \in X$, we have $x\succsim y$, $y\succsim x$, or both.

$(.5,.5)\not \succsim (.5,.5)$, thus the preference relation is not complete.

## Transitivity

> A preference relation is transitive, if $x\succsim y$ and $y \succsim z$ imply $x\succsim z$.

$(1,0)\succsim (.5,.5)$ and $(.5,.5) \succsim (0,1)$ hold but $(1,0)\not \succsim (0,1)$, thus the preference relation is not transitive.

## Continuity

> A preference relation is continuous if for all sequences ${(x\_i,y\_i)}\_{i=1}^{\infty}$ converging to $(x,y)$ with $\forall i: x\_i \succsim y\_i$ we have $x \succsim y$.

The preference relation does not violate continuity. Consider a sequence $x\_i \succsim y\_i$ which converges to $x,y$. These sequences can only be such that $x\_i=x$ and $y\_i=y$, and $x\neq y$, since all other $x\_i,y\_i$ either do not converge to $x,y$, or do not fulfill $x\_i \succsim y\_i$. But clearly if $x\_i\succsim y\_i$ then $x\succsim y$.

## Monotonicity

> A preference relation is monotone, if $x \geq y$ implies $x \succsim y$.

The $\geq$ relation considers all elements of $X$ incomparable, thus the preference relation is monotone.

Thus, we have an intransitive, incomplete, monotone, continuous preference relation.

> **Comments:**
> **username_1**: I assume that $x \leq 1, y \leq 1$, but even so the definition of your relation seems incomplete. What is preferred, (0.1,0.9) or (0,1)? (And what about the other pairings?) By the $\succeq$ between (0.5,0.5) and (0,1) do you mean $\sim$?  
> **username_3**: Thank you for pointing out the typing error. Regarding the remaining comments about providing an incomplete relation: This is exactly the point. We are looking for a preference relation which is intransitive + incomplete, but at the same time monotone and continuous. If we start out with a preference relation that is complete, this would defeat the purpose.  
> **username_1**: I see. So you mean the relation is only defined exactly where you defined it. This is not always the case. E.g.: 3 < 5, but the relation also where I have not defined it.  
> **username_3**: a relation is always only "defined where one defines it". Formally, a relation is a subset of the cartesian product of a set. For the definition of the relation, the specification of that subset is sufficient. Thus, you can define the relation < on the real numbers such that 3<5. This will not correspond to the usual definition, but it is nonetheless a valid specification of an (incomplete) relation.  
> **username_1**: Alright I'll rephrase my comment: I have tought you merely gave some  examples about how your relation would work and not an exact definition but now I understand what you meant.  
> **username_3**: ok, i hope now it is clear  

---

### Answer by username_4 on 2015-09-20. Score: 2

Transitivity of preferences appeals to some "intuitive" notion of "consistency of human mind" and it can be argued that any exceptions are the "exceptions to the *rule*", and so we *do* have an adequate abstract rule.

In comparison, Completeness is much more of a "leap of faith". It hangs in the air, stemming from nothing, related to nothing (**so the answer to your question is no**). Perhaps it can be supported by some vulgar remark that "if you *press* a person enough, then he will eventually order any pair you will put in front of him, even if only to get rid of you", but of course this, while looking good in practice, will never work in theory.

So we just define Completeness to exist... why? In order to avoid rather *unmanageable* issues down the road. How useful will it be to work with non-complete preferences? How useful would it be to say "I have this model, it may result, it may not, depending on whether preferences are complete or not"... what's the use of it? We would then be forced to come up with an *alternative* decision rule: "Assuming that preferences are not complete, then if the person encounters a pair that he cannot order..." -he does *what*? Flip a coin? But this would make "incompleteness" equivalent to indifference...

What else? This line of thought may be very stimulating, but it is also very challenging, and certainly path-breaking, if indeed, such a path exists or can be created. (In my opinion, various theoretical explorations of the "fuzzy" variety try to find a "middle way" for exactly this problem -where they consider a situation where the person neither has complete preferences, nor is completely "frozen" when a "difficult" pair comes up).


---

