Equivalence of the definitions of the Axiom of Continuity

I am trying to understand this topic in more detail. Below are the definitions of the Axiom of Continuity given by different authors. Defn. 1 is given by Ariel Rubinstein and I have seen lecture notes using Defn. 2.

Def. 1: A preference relation $$(\succeq)$$ on $$X$$ is continuous if for $$r \succ s$$, there exists balls $$B_r$$ and $$B_s$$ around $$r$$ and $$s$$ (respectively) such that for every $$x$$ in $$B_r$$ and $$y$$ in $$B_s$$, we have $$x \succ y$$.

Def. 2: $$\text{(Part 1)}$$ If $$x_n \rightarrow x$$ and $$x_n \succeq y$$ $$\forall \ n$$ then $$x \succeq y$$, and $$\text{(Part 2)}$$ if $$x_n \preceq y$$ $$\forall \ n$$ then $$x \preceq y$$.

My questions:

1. I have seen Defn. 1 and Defn. 2 in microeconomic books. I do not find them to be equal; in fact, Defn. 2 is the converse of Defn. 1. Isn't it so?

2. Many authors simply write that "if A is more preferable to B and B is 'sufficiently close' to C, then A is preferable to C". If I am to go through Def. 2 and write it in a more formal way, then can Part 1 and 2 of Def. 2 correspond to (or be re-written as) Part 1 and 2 (respectively) as in the following new Def. 3: $$\text{(Part 1) } (A \succeq B) \land (A \rightarrow C) \implies C \succeq B \\ \text{(Part 2) } (A \succeq B) \land (B \rightarrow C) \implies A \succeq C$$

• Rubinstein's lecture note actually have a proof of the equivalence of the two definitions. Jun 24 '21 at 14:26

Consider the following 4 conditions.

1. If $$x \succ y$$ there is a ball $$B_r$$ around $$x$$ such that for all $$z \in B_r$$, $$z \succ y$$.
2. If $$y \succ x$$ there is a ball $$B_s$$ around $$x$$ such that for all $$z \in B_s$$ we have $$y \succ z$$.
3. if $$x_n \to x$$ and $$x_n \succeq y$$ then $$x \succeq y$$.
4. 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.

Other continuity conditions

Consider in addition the following 2 conditions

1. 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 $$w \in B_s$$, $$z \succ w$$.

2. 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:

1. $$\forall x, y$$, 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

2. $$\forall x, y$$ 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. 1. There is an $$N_3$$ such that for all $$n \ge N_3$$: $$x_n \succeq x.$$ 2. 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. • Do you mean Defn. 2 and Defn. 3 are equivalent? I think you said exactly the opposite the other day and that has still kept me confused. (Because I find Def. 2, 3 to be equivalent.) Jun 24 '21 at 21:26 • I do not as well see how$B_r \ni x \succ y \in B_s$. Your answer probably just shows$x \succ r$and$y \prec s\$. Jun 26 '21 at 17:26
• In fact, I don't see how exactly the statements 1 and 2 combined is equivalent to Defn. 1 of mine. I don't understand the double limit case there. Jun 26 '21 at 18:50
• @dictatemetokcus Sorry, but I also have a life... and sometimes other stuff to do, which means that I can't continuously check and respond to queries. If you don't like my answer, just downvote. I completed your question and fixed the typo in 1. I also had some looking up to do as the argument is not that simple. I hope you're happy now. (?)
– tdm
Jun 27 '21 at 13:38
• @dictatemetokcus btw, if you want me to delete the answer, just let me know (no problem).
– tdm
Jun 27 '21 at 14:43