Consider the following 4 conditions.
- If $x \succ y$ there is a ball $B_r$ around $x$ such that for all $z \in B_r$, $z \succ y$.
- If $y \succ x$ there is a ball $B_s$ around $x$ such that for all $z \in B_s$ we have $y \succ z$.
- if $x_n \to x$ and $x_n \succeq y$ then $x \succeq y$.
- If $x_n \to x$ and $y \succeq x_n$ then $y \succeq x$.
Consider the sets
$$
\begin{align*}
&UC(y) = \{x: x \succeq y\},\\
&LC(y) = \{x: y \succeq x\},\\
&SUC(y) = \{x: x \succ y\},\\
&SLC(y) = \{x: y \succ x\}.
\end{align*}
$$
We have the following equivalences:
- 1 is equivalent to the assumption that $SUC(y)$ is open.
- 2 is equivalent to the assumption that $SLC(y)$ is open
- 3 is equivalent to the assumption that $UC(y)$ is closed
- 4 is equivalent to the assumption that $LC(y)$ is closed.
We also have that:
- $SUC(y)$ is the complement of $LC(y)$.
- $SLC(y)$ is the complement of $UC(y)$.
We also know that a set is open if and only if its complement is closed. From this it follows that $1$ is equivalent to $4$ and $2$ is equivalent to $3$.
Alternatively you can also proof it directly
Th 1 is equivalent to 4
proof ($1 \to 4$) Let $x_n \to x$ and $y \succeq x_n$. Towards a contradiction assume that $y \not \succeq x$ which means that $x \succ y$. Then we know that there exists an open ball $B_r$ around $x$ such that for all $z \in B_r$, $z \succ y$. But for $n$ large enough we have that $x_n \in B_r$, so for $n$ large enough also $x_n \succ y$, which is a contradiction.
$(4 \to 1)$. Assume that $x \succ y$ and assume that for all open balls $B_r$ around $x$ there exists an $z_r$ such that $z_r \not \succ y$, which means that $y \succeq z_r$. Take a sequence $r_n \to 0$ which generates a sequence $z_n$ with $y \succeq z_n$ and $z_n \to x$. Assumption 4 then gives that $y \succeq x$, which gives the desired contradiction.
The equivalence between 2 and 3 can be shown in a similar way.
... continue answer
Other continuity conditions
Consider in addition the following 2 conditions
If $x \succ y$ there is a ball $B_r$ around $x$ and a ball $B_s$ around $y$ such that for all $z \in B_r$ and $w \in B_s$, $z \succ w$.
If $x_n \to x$ and $y_n \to y$ and for all $n$, $y_n \succeq x_n$ then $y \succeq x$.
Let us first show that these two are equivalent
Th $5$ is equivalent to $6$.
proof Assume that $5$ holds. Let $x_n \to x$ and $y_n \to y$ with $y_n \succeq x_n$ for all $n$. Towards a contradiction assume that $y \not \succeq x$. Then $x \succ y$. As such, there should be a ball $B_r$ around $x$ and $B_s$ around $y$ such that for all $z \in B_r$ and $w \in B_s$, $z \succ w$. Now, for all $n$ big enough $x_n \in B_r$ and $y_n \in B_s$. As such, for $n$ big enough $x_n \succ y_n$, a contradiction.
For the reverse, assume that $6$ holds $x \succ y$ and, towards a contradiction for all balls $B_r$ around $x$ and $B_s$ around $y$, there are $z \in B_r$ and $w \in B_s$ such that $w \succeq z$. Let $r_n \to 0$ and $s_n \to 0$ then we can generate a sequence $z_n \in B_{r_n}$ and $w_n \in B_{s_n}$ such that $z_n \to x$, $w_n \to y$. And for all $n$, $w_n \succeq z_n$. this gives (by $6$) that $y \succeq x$, a contradiction.
Connecting the various conditions together
Remember from the first part the following two conditions:
$\forall x, y$, if $x_n \to x$ and $y \succeq x_n$ for all $n$ then $y \succeq x$ $\leftrightarrow$ $\forall y$, $LC(y)$ is closed, $\leftrightarrow$ $\forall y$, $SUC(y)$ is open
$\forall x, y$ if $y_n \to y$ and $y_n \succeq x$ for all $n$ then $y \succeq x$ $\leftrightarrow$ $\forall x$, $UC(x)$ is closed $\leftrightarrow$ $\forall x$, $SLC(x)$ is open
The following result shows the equivalence between $5$ (or $6$)$ and $7$ and $8$. It is also exercise 3.C.3 in MWG (if you would be interested). The proof is not so self-evident.
Th $5$ (or equivalently $6$) hold if and only if both $7$ and $8$ hold.
proof That $5$ (or $6$) imply $7$ and $8$ is obvious as we can take the constant sequence. For the reverse, assume that $7$ and $8$ hold and that $6$ is not true. This means that $x_n \to x$, $y_n \to y$, $y_n \succeq x_n$ for all $n$ and $x \succ y$. Then as $SUC(y)$ is open, there is an $N_1$ such that for all $n \ge N_1$:
$$
x_n \succ y.
$$
As $SLC(x)$ is open, there is also an $N_2$ such that for all $n \ge N_2$:
$$
x \succ y_n.
$$
There are two possible cases.
There is an $N_3$ such that for all $n \ge N_3$:
$$
x_n \succeq x.
$$
There is a subsequence $k(n)$ such that for all $n$
$$
x \succ x_{k(n)}.
$$
If $1.$ is the case, then for all $n \ge \max\{N_3, N_2\}$,
$$
x_n \succeq x \succ y_n,
$$
a contradiction.
If $2.$ is the case, then we can find an $m$ such that $k(m) \ge N_1$. Then:
$$
x \succ x_{k(m)} \succ y
$$
As $SUC(x_{k(m)})$ is open and $x_n \to x$, we know there is an $N_4$ such that for all $n \ge N_4$:
$$
y_n \succeq x_n \succ x_{k(m)} \succ y
$$
Taking the limit for $n \to \infty$ we see that:
$$
y \succeq x_{k(m)} \succ y.
$$
again a contradiction.
