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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.

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    $\begingroup$ 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). $\endgroup$ Commented Feb 3, 2022 at 10:19
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    $\begingroup$ Let $y=x$, then Def. 2 implies $z≽x$ for all $z$ sufficiently close to $x$. So Def. 2 is obviously wrong. $\endgroup$
    – VARulle
    Commented Feb 3, 2022 at 11:10
  • $\begingroup$ Definition 1 is equivalent to Definition 2 if one replaces the weak preference $\succeq$ by the strict preference $\succ$ in Definition 2. $\endgroup$ Commented Feb 3, 2022 at 13:02
  • $\begingroup$ @VARulle Thanks for pointing it out! I have added the correct definition (thanks to user tdm). $\endgroup$ Commented Feb 3, 2022 at 14:12
  • $\begingroup$ @MichaelGreinecker Can you tell me how they are equivalent? I still could not figure it out. $\endgroup$ Commented Feb 3, 2022 at 14:12

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Feb 3, 2022 at 14:15
  • $\begingroup$ 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. $\endgroup$
    – tdm
    Commented Feb 3, 2022 at 18:49
  • $\begingroup$ I see. Can you provide a counter-example in that case? $\endgroup$ Commented Feb 4, 2022 at 5:56
  • $\begingroup$ Take any utility function that is upper semi-continuous but not lower semi-continuous or vice versa. $\endgroup$
    – tdm
    Commented Feb 4, 2022 at 9:06
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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Feb 4, 2022 at 11:26
  • $\begingroup$ It does (see my updated answer). In your argument, you seem to work with $\succeq$ when you really should use $\succ$. $\endgroup$ Commented Feb 4, 2022 at 11:38
  • $\begingroup$ 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.) $\endgroup$ Commented Feb 4, 2022 at 16:27
  • $\begingroup$ @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. $\endgroup$ Commented Feb 4, 2022 at 17:21
  • $\begingroup$ "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! $\endgroup$ Commented Feb 4, 2022 at 19:19

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