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: $$\text{(Part 1) } (A \succeq B) \land (A \rightarrow C) \implies C \succeq B \\ \text{(Part 2) } (A \succeq B) \land (B \rightarrow C) \implies A \succeq C$$


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.

... continue answer

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.

  • $\begingroup$ 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.) $\endgroup$ Jun 24 '21 at 21:26
  • $\begingroup$ 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 $y \prec s$. $\endgroup$ Jun 26 '21 at 17:26
  • $\begingroup$ 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. $\endgroup$ Jun 26 '21 at 18:50
  • $\begingroup$ @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. (?) $\endgroup$
    – tdm
    Jun 27 '21 at 13:38
  • $\begingroup$ @dictatemetokcus btw, if you want me to delete the answer, just let me know (no problem). $\endgroup$
    – tdm
    Jun 27 '21 at 14:43

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