Is Varian's definition of continuity of preference equivalent to standard definitions?

Question

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.

Answer 1

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.

Answer 2

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.