<p>Here is how one can show that Definition 1 implies Definition 2. We do the contrapositive, we show that if Definition 2 fails then Definition 1 will fail too.</p>
<p>Suppose that <span class="math-container">$x\succ y$</span>, but for every <span class="math-container">$\epsilon&gt;0$</span>, there exist <span class="math-container">$x'$</span> and <span class="math-container">$y'$</span> such that <span class="math-container">$x'\in B(x,\epsilon)$</span> and <span class="math-container">$y'\in B(x,\epsilon)$</span> but <span class="math-container">$y'\succeq x'$</span>.</p>
<p>Then we can find for each positive natural number points <span class="math-container">$x_n$</span> and <span class="math-container">$y_n$</span> such that <span class="math-container">$x_n\in B(x,1/n)$</span>, <span class="math-container">$y_n\in B(y,1/n)$</span>, and <span class="math-container">$y_n\succeq x_n$</span>. Then <span class="math-container">$(x_n)\to x$</span>, <span class="math-container">$(y_n)\to y$</span>, <span class="math-container">$y_n\succeq x_n$</span> for all <span class="math-container">$n$</span>, but not <span class="math-container">$x\succeq y$</span>, which contradicts Definition 1.</p>

CC BY-SA 4.0
2022-02-04T11:26:08.727Z
Does Def. 2 $\implies$ 1? If not, where did I go wrong? The other answer rules this out, but since you wrote (as a comment) that the two definitions are equivalent, could you please clarify?
not tdm's twin
CC BY-SA 4.0
2022-02-04T11:38:15.440Z
It does (see my updated answer). In your argument, you seem to work with $\succeq$ when you really should use $\succ$.
Michael Greinecker
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2022-02-04T16:27:06.423Z
Consider the $X = \mathbb{R}$ the set of consumption bundles and utility function $U : X \to \{-1,1\}$ and .
\begin{equation}
  U(x) =
    \begin{cases}
      1 & \text{if } x \geq 0\\
      -1 & \text{if } x < 0
    \end{cases}       
\end{equation}
See that Definition 1 (standard) holds here but Varian's definition fails. To show that the latter fails, choose $(x,y,z) = (0, x - \epsilon, x + \epsilon)$ for some "small" $\epsilon > 0$. Isn't it so? (I took user tdm, the other answer's hints.)
not tdm's twin
CC BY-SA 4.0
2022-02-04T17:21:29.940Z
@Gang'sBigBoss You are right; I just realized that this is only one direction. To get full equivalence, you need both the strictly better and strictly worse sets to be open. Though Varian only writes that the former is a consequence of his official definition.
Michael Greinecker
CC BY-SA 4.0
2022-02-04T19:19:47.193Z
`"Though Varian only writes that the former is a consequence of his official definition."` 

Well in that case, Varian is incorrect, right? The converse is true and this is where we stand I believe. Thanks for the help!
not tdm's twin
CC BY-SA 4.0
2022-02-04T21:19:48.517Z
Varian's official definition is that the weakly-better sets and the weakly-worse sets are open. He then states that this implies what you call Definition 2. It does.
Michael Greinecker
CC BY-SA 4.0
2022-02-05T00:19:06.640Z
That makes sense, thanks again.
not tdm's twin
CC BY-SA 4.0
2022-02-03T14:47:31.350Z
2022-02-04T17:20:08.200Z
2022-02-04T17:20:08.200Z
Michael Greinecker
<p>What Varian (Microeconomic Analysis, p 95) says is that:</p>
<blockquote>
<p>If <span class="math-container">$x$</span> is strictly preferred to <span class="math-container">$y$</span> and if <span class="math-container">$z$</span> is a bundle that is close enough to <span class="math-container">$x$</span> then <span class="math-container">$z$</span> must be strictly preferred to <span class="math-container">$y$</span>.</p>
</blockquote>
<p>This is a consequence of the standard definition. Indeed, if we formalize this, it states that:</p>
<ul>
<li>If <span class="math-container">$x \succ y$</span> and if <span class="math-container">$z$</span> is close enough to <span class="math-container">$x$</span> then <span class="math-container">$z \succ y$</span>.</li>
<li>Equivalently, if <span class="math-container">$x \succ y$</span> then there is a <span class="math-container">$\varepsilon &gt; 0$</span> such that for all <span class="math-container">$z \in B(x, \varepsilon)$</span>, <span class="math-container">$z \succ y$</span>.</li>
<li>Equivalently, every <span class="math-container">$x$</span> in <span class="math-container">$\{w| w \succ y\}$</span> is an interior point.</li>
<li>Equivalently, <span class="math-container">$\{w| w \succ y\}$</span> is an open set.</li>
<li>Equivalently, as preferences are complete, <span class="math-container">$\{w| y \succeq w\}$</span> is a closed set.</li>
</ul>
<p>The latter follows from continuity as the following is true:</p>
<p><strong>Proposition</strong> If preferences are continuous (as in Definition 1) then for all <span class="math-container">$y$</span> the sets <span class="math-container">$\{w| y \succeq w\}$</span> are closed.</p>
<p><em>Proof:</em> Assume preferences are continuous. Let consider a convergent sequence <span class="math-container">$(w_n)$</span> in <span class="math-container">$\{w| y \succeq w\}$</span> then for all <span class="math-container">$n$</span>, <span class="math-container">$y \succeq w_n$</span>. So if we define the sequence <span class="math-container">$(y_n)$</span> with <span class="math-container">$y_n = y$</span> for all <span class="math-container">$n$</span> we have that <span class="math-container">$y_n \succeq w_n$</span> for all <span class="math-container">$n$</span>. As <span class="math-container">$w_n \to w$</span> and <span class="math-container">$y_n \to y$</span> we have that <span class="math-container">$y \succeq w$</span>. As such, <span class="math-container">$\{w| y \succeq w\}$</span> contains all its limit points, i.e. it is a closed set.</p>

CC BY-SA 4.0
2022-02-03T14:15:15.383Z
Thanks for mentioning the correct definition. But I am not sure if this proves the equivalence of the two definitions (since you have not proven the equivalence of the statements you've written). Can you show that Def. 1 implies Def. 2? Ideally, I am looking for a short, simple proof rather than a sequence of equivalence proofs.
not tdm's twin
CC BY-SA 4.0
2022-02-03T18:49:26.713Z
Well they are not equivalent as continuity requires both $\{w| w \succ y\}$ and $\{w| y \succ w\}$ to be open sets, not just one of them.
tdm
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2022-02-04T05:56:07.570Z
I see. Can you provide a counter-example in that case?
not tdm's twin
CC BY-SA 4.0
2022-02-04T09:06:52.380Z
Take any utility function that is upper semi-continuous but not lower semi-continuous or vice versa.
tdm
CC BY-SA 4.0
2022-02-03T12:22:37.930Z
2022-02-03T12:22:37.930Z
tdm
<p>Here are two definitions of continuity of preferences. Denote the (weak) preference relation by ≽. We assume completeness, reflexivity and transitivity. <strong>Assume non-satiation or strict monotonicity only if necessary (and if you do so, please mention).</strong></p>
<p><strong>Definition 1 (Standard):</strong> If <span class="math-container">$(x_n)$</span> and <span class="math-container">$(y_n)$</span> are two sequences such that <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y_n \to y$</span>, then if <span class="math-container">$x_n ≽ y_n$</span> for all <span class="math-container">$n$</span>, we have <span class="math-container">$x ≽ y$</span>.</p>
<p><strong>Definition 2 (Varian/NS):</strong> If <span class="math-container">$x \succ y$</span> and <span class="math-container">$z$</span> is &quot;sufficiently close&quot; to <span class="math-container">$x$</span>, then <span class="math-container">$z \succ y$</span>.</p>
<p><strong>Can we prove that these two are equivalent?</strong></p>
<p>Here's an attempt.</p>
<p>Proof that Def. 2 implies Def. 1: Suppose not. Assume Varian's definition. Then if we have <span class="math-container">$(x_n)$</span> and <span class="math-container">$(y_n)$</span> with the given criteria, the result would be <span class="math-container">$y ≽ x$</span> (due to completeness). Since (for all <span class="math-container">$n$</span>) <span class="math-container">$y_n$</span> is sufficiently close to <span class="math-container">$y$</span>, by Def. 2, <span class="math-container">$y_n ≽ y$</span> which is not necessarily true in general. Thus, Def. 2 implies Def. 1.</p>
<p><strong>Can we show that Def. 1 implies Def. 2? It seems like we need additional conditions, but I can't figure out what all is needed, if at all.</strong></p>
<p>Note: &quot;Sufficiently close&quot; is an informal terminology used by Varian, so you can treat it this way - if <span class="math-container">$x$</span> and <span class="math-container">$z$</span> are sufficiently close, that is, if <span class="math-container">$z$</span> lies in the <span class="math-container">$\epsilon$</span>-ball of <span class="math-container">$x$</span> (written <span class="math-container">$B(x,\epsilon)$</span>), then <span class="math-container">$\epsilon &gt; 0$</span> can be set as small as <strong>you</strong> wish.</p>
<p><strong>Edit:</strong> Varian/Nicholson_Snyder used strict preferences which has now been incorporated.</p>

CC BY-SA 4.0
2022-02-03T10:19:13.737Z
Can you give a reference to where Varian is supposed to have given this definition? In his book "Microeconomic Analysis," he gives a different definition. I did not find any definition in his book "Intermediate Microeconomics" (7th ed).
Michael Greinecker
CC BY-SA 4.0
2022-02-03T11:10:12.163Z
Let $y=x$, then Def. 2 implies $z≽x$ for all $z$ sufficiently close to $x$. So Def. 2 is obviously wrong.
VARulle
CC BY-SA 4.0
2022-02-03T13:02:12.310Z
Definition 1 is equivalent to Definition 2 if one replaces the weak preference $\succeq$ by the strict preference $\succ$ in Definition 2.
Michael Greinecker
CC BY-SA 4.0
2022-02-03T14:12:00.357Z
@VARulle Thanks for pointing it out! I have added the correct definition (thanks to user tdm).
not tdm's twin
CC BY-SA 4.0
2022-02-03T14:12:11.640Z
@MichaelGreinecker Can you tell me how they are equivalent? I still could not figure it out.
not tdm's twin
<microeconomics><mathematical-economics><consumer-theory><preferences><proof>
Is Varian's definition of continuity of preference equivalent to standard definitions?
not tdm's twin
# Is Varian's definition of continuity of preference equivalent to standard definitions?

(Asked by: username_0 on 2022-02-03)

Here are two definitions of continuity of preferences. Denote the (weak) preference relation by ≽. We assume completeness, reflexivity and transitivity. **Assume non-satiation or strict monotonicity only if necessary (and if you do so, please mention).**

**Definition 1 (Standard):** If $(x\_n)$ and $(y\_n)$ are two sequences such that $x\_n \to x$ and $y\_n \to y$, then if $x\_n ≽ y\_n$ for all $n$, we have $x ≽ y$.

**Definition 2 (Varian/NS):** If $x \succ y$ and $z$ is "sufficiently close" to $x$, then $z \succ y$.

**Can we prove that these two are equivalent?**

Here's an attempt.

Proof that Def. 2 implies Def. 1: Suppose not. Assume Varian's definition. Then if we have $(x\_n)$ and $(y\_n)$ with the given criteria, the result would be $y ≽ x$ (due to completeness). Since (for all $n$) $y\_n$ is sufficiently close to $y$, by Def. 2, $y\_n ≽ y$ which is not necessarily true in general. Thus, Def. 2 implies Def. 1.

**Can we show that Def. 1 implies Def. 2? It seems like we need additional conditions, but I can't figure out what all is needed, if at all.**

Note: "Sufficiently close" is an informal terminology used by Varian, so you can treat it this way - if $x$ and $z$ are sufficiently close, that is, if $z$ lies in the $\epsilon$-ball of $x$ (written $B(x,\epsilon)$), then $\epsilon > 0$ can be set as small as **you** wish.

**Edit:** Varian/Nicholson\_Snyder used strict preferences which has now been incorporated.

**Tags:** microeconomics, mathematical-economics, consumer-theory, preferences, proof

> **Comments:**
> **username_1**: Can you give a reference to where Varian is supposed to have given this definition? In his book "Microeconomic Analysis," he gives a different definition. I did not find any definition in his book "Intermediate Microeconomics" (7th ed).  
> **username_2**: Let $y=x$, then Def. 2 implies $z≽x$ for all $z$ sufficiently close to $x$. So Def. 2 is obviously wrong.  
> **username_1**: Definition 1 is equivalent to Definition 2 if one replaces the weak preference $\succeq$ by the strict preference $\succ$ in Definition 2.  
> **username_0**: @username_2 Thanks for pointing it out! I have added the correct definition (thanks to user username_3).  
> **username_0**: @MichaelGreinecker Can you tell me how they are equivalent? I still could not figure it out.  

---

## 2 answers

### Answer by username_3 on 2022-02-03. Score: 6 | Accepted answer

What Varian (Microeconomic Analysis, p 95) says is that:

> If $x$ is strictly preferred to $y$ and if $z$ is a bundle that is close enough to $x$ then $z$ must be strictly preferred to $y$.

This is a consequence of the standard definition. Indeed, if we formalize this, it states that:

* If $x \succ y$ and if $z$ is close enough to $x$ then $z \succ y$.
* Equivalently, if $x \succ y$ then there is a $\varepsilon > 0$ such that for all $z \in B(x, \varepsilon)$, $z \succ y$.
* Equivalently, every $x$ in $\{w| w \succ y\}$ is an interior point.
* Equivalently, $\{w| w \succ y\}$ is an open set.
* Equivalently, as preferences are complete, $\{w| y \succeq w\}$ is a closed set.

The latter follows from continuity as the following is true:

**Proposition** If preferences are continuous (as in Definition 1) then for all $y$ the sets $\{w| y \succeq w\}$ are closed.

*Proof:* Assume preferences are continuous. Let consider a convergent sequence $(w\_n)$ in $\{w| y \succeq w\}$ then for all $n$, $y \succeq w\_n$. So if we define the sequence $(y\_n)$ with $y\_n = y$ for all $n$ we have that $y\_n \succeq w\_n$ for all $n$. As $w\_n \to w$ and $y\_n \to y$ we have that $y \succeq w$. As such, $\{w| y \succeq w\}$ contains all its limit points, i.e. it is a closed set.

> **Comments:**
> **username_0**: Thanks for mentioning the correct definition. But I am not sure if this proves the equivalence of the two definitions (since you have not proven the equivalence of the statements you've written). Can you show that Def. 1 implies Def. 2? Ideally, I am looking for a short, simple proof rather than a sequence of equivalence proofs.  
> **username_3**: Well they are not equivalent as continuity requires both $\{w| w \succ y\}$ and $\{w| y \succ w\}$ to be open sets, not just one of them.  
> **username_0**: I see. Can you provide a counter-example in that case?  
> **username_3**: Take any utility function that is upper semi-continuous but not lower semi-continuous or vice versa.  

---

### Answer by username_1 on 2022-02-03. Score: 6

Here is how one can show that Definition 1 implies Definition 2. We do the contrapositive, we show that if Definition 2 fails then Definition 1 will fail too.

Suppose that $x\succ y$, but for every $\epsilon>0$, there exist $x'$ and $y'$ such that $x'\in B(x,\epsilon)$ and $y'\in B(x,\epsilon)$ but $y'\succeq x'$.

Then we can find for each positive natural number points $x\_n$ and $y\_n$ such that $x\_n\in B(x,1/n)$, $y\_n\in B(y,1/n)$, and $y\_n\succeq x\_n$. Then $(x\_n)\to x$, $(y\_n)\to y$, $y\_n\succeq x\_n$ for all $n$, but not $x\succeq y$, which contradicts Definition 1.

> **Comments:**
> **username_0**: Does Def. 2 $\implies$ 1? If not, where did I go wrong? The other answer rules this out, but since you wrote (as a comment) that the two definitions are equivalent, could you please clarify?  
> **username_1**: It does (see my updated answer). In your argument, you seem to work with $\succeq$ when you really should use $\succ$.  
> **username_0**: Consider the $X = \mathbb{R}$ the set of consumption bundles and utility function $U : X \to \{-1,1\}$ and . \begin{equation}   U(x) =     \begin{cases}       1 & \text{if } x \geq 0\\       -1 & \text{if } x < 0     \end{cases}        \end{equation} See that Definition 1 (standard) holds here but Varian's definition fails. To show that the latter fails, choose $(x,y,z) = (0, x - \epsilon, x + \epsilon)$ for some "small" $\epsilon > 0$. Isn't it so? (I took user username_3, the other answer's hints.)  
> **username_1**: @Gang'sBigBoss You are right; I just realized that this is only one direction. To get full equivalence, you need both the strictly better and strictly worse sets to be open. Though Varian only writes that the former is a consequence of his official definition.  
> **username_0**: `"Though Varian only writes that the former is a consequence of his official definition."`   Well in that case, Varian is incorrect, right? The converse is true and this is where we stand I believe. Thanks for the help!  
> **username_1**: Varian's official definition is that the weakly-better sets and the weakly-worse sets are open. He then states that this implies what you call Definition 2. It does.  
> **username_0**: That makes sense, thanks again.  

---

