We have two definitions of the continuity of preferences:

Def 1: $\succcurlyeq$ is continuous if for any sequences $\{x^n\} \subset X$ and $\{y^n\} \subset X$, then $n \in \mathbb{N}$ such that,

  • $\forall n, \quad x^n \succcurlyeq y^n$
  • $\lim_{n \rightarrow \infty} x^n = x, \quad \lim_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$

then $x \succcurlyeq y$


Def 2: $\succcurlyeq$ is continuous if whenever $x \succ y$, $\exists \ B_x, \ B_y$, open balls around $x, y$, such that $\forall \ x' \in B_x, \ y' \in B_y$, then we have $x' \succ y'$.

Show that the following definitions are equivalent to each other.


1 Answer 1


We want to show that for $\succcurlyeq$ on $X$, Def 1 $\iff$ Def 2

$\boxed \Longrightarrow$

Assume that $\succcurlyeq$ is continuous by Def 1.

Let us say $x \succ y$. Denote our open-balls as $B(x, r)$, an open ball around $x$ of radius $r$. Suppose $\forall n, \ \exists \ x^n \in B(x, \frac{1}{n}), \ y^n \in B(y, \frac{1}{n})$ such that $y^n \succcurlyeq x^n$. But then we have constructed $\{x^n\} \rightarrow x$ and $\{y^n\} \rightarrow y$, and by Def 1, $y \succcurlyeq x$, which is a contradiction.

$\boxed \Longleftarrow$

Assume that $\succcurlyeq$ is continuous by Def 2.

Let us take sequences ${x^n} \subset X$ and ${y^n} \subset X$ where $\forall n, \quad x^n \succcurlyeq y^n$ and $\lim_{n \rightarrow \infty} x^n = x, \quad \lim_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$ for $n \in \mathbb{N}$,

BUT $x \succcurlyeq y$ is false instead of true. We want to show this leads to a contradiction.

If $x \succcurlyeq y$ is false, (so $y \succ x$) then $\exists B_x, B_y$ such that $\forall y' \in B_y, x' \in B_x$, we have $y' \succ x'$. Because $\{x^n\} \rightarrow x, \{y^n\} \rightarrow y$, there exists $N$ large enough such that $\forall n>N$, we have $y^n \in B_y, x^n \in B_x$.

Thus $\forall n > N$, we have $y^n \succ x^n$, which contradicts $\forall n, \ x^n \succcurlyeq y^n$.


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