Equivalence of definitions of the Axiom of Continuity in Economics Statement $1$: If $(a_n, b_n)_{n \geq 1}^{\infty}$ is a sequence in $X \times X$ satisfying $a_n \succsim b_n$ and $\displaystyle\lim_{n \to \infty} (a_n, b_n) = (a,b)$, then $a \succsim b$.
Statement $2$: If $x_n \to x$ and $x_n \succsim y \ \forall \ n \in \mathbb{N}$, then $x \succsim y$ and if $x_n \precsim y \ \forall \ n \in \mathbb{N}$, then $x \precsim y$.
Definition: $\succsim$ is a binary relation on $X \times X$ such that it is complete ($x \succ y, y \succ x, x \sim y$ are the only possibilities) and transitive. We denote $x$ is more preferable to $y$ by $x \succsim y$. Here, $\succ$ is strict, $\succsim$ is weak and $\sim$ denotes indifference (that is, $x$ is as preferable as $y$ to the consumer).
Statement $1$ is essentially the definition of the Axiom of Continuity in Economics. $X$ is supposed to be the choice set, an infinite subset of the Euclidean space. I am trying to verify if the second statement means the same as the other one.
$1 \implies 2:$ Consider the sequence $(x_n)$ converging to $x$ such that for each $n \in \mathbb{N}$, we have $x_n \succsim y$. In $(1)$, define $b_{n \geq 1} := y$ and we have the desired result. Similarly, do it the other way and we have the second part of $(2)$ proved.
However, the problem lies in proving $2 \implies 1$. I have been trying this for quite sometime and I haven't made any progress yet. I did not find much help on the Economics Stack site either.
 A: The two statements are indeed equivalent to each other, and either one can be used as the definition of a continuous preference on the set $X$.
Assume that Statement 2 is true, then ${\rm UC}(y)=\{x : x \succsim y\}$ and ${\rm LC}(y)=\{x : x \precsim y\}$ are closed sets for each $y$. Since the preference $\succsim$ is complete, ${\rm SUC}(y) = \{x : x \succ y\}$ and ${\rm SLC}(y) = \{x : x\prec y\}$ are open sets for each $y$.
Assume that Statement 1 is not true. Then there is a sequence $(a_n,b_n)\in X\times X$ such that $a_n\succsim b_n$ for each $n$, $a_n \to a$ and $b_n \to b$ as $n\to\infty$, but $a\prec b$.
The first step is to choose some particular $n$ such that $b\succ a_n \succsim b_n \succ a$. Choose a neighborhood $U$ of $a$ with $U\subset {\rm SLC}(b)$, and choose a neighborhood $V$ of $b$ with $V\subset {\rm SUC}(a)$. Since $a_n \to a$ and $b_n \to b$, we must have $a_n \in U$ and $b_n \in V$ for $n$ large enough. Pick one such $n$, then
$$b\succ a_n \succsim b_n \succ a$$
Keep $n$ fixed, and choose a smaller neighborhood $U_0$ of $a$ with $U_0\subset {\rm SLC}(b_n)$, and a smaller neighborhood $V_0$ of $b$ with $V_0 \subset {\rm SUC}(a_n)$. Again, if $m$ is large enough, $a_m \in U_0$ and $b_m \in V_0$. Pick one such $m$, then $a_m\prec b_n$ and $b_m\succ a_n$. But together this yields
$$b_m\succ a_n \succsim b_n \succ a_m$$
which contradicts the requirement $a_m \succsim b_m$. Hence, if Statement 2 is true, then Statement 1 must also be true.
