<p>Consider the following 4 conditions.</p>
<ol>
<li>If  <span class="math-container">$x \succ y$</span> there is a ball <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> such that for all <span class="math-container">$z \in B_r$</span>, <span class="math-container">$z \succ y$</span>.</li>
<li>If <span class="math-container">$y \succ x$</span> there is a ball <span class="math-container">$B_s$</span> around <span class="math-container">$x$</span> such that for all <span class="math-container">$z \in B_s$</span> we have <span class="math-container">$y \succ z$</span>.</li>
<li>if <span class="math-container">$x_n \to x$</span> and <span class="math-container">$x_n \succeq y$</span> then <span class="math-container">$x \succeq y$</span>.</li>
<li>If <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y \succeq x_n$</span> then <span class="math-container">$y \succeq x$</span>.</li>
</ol>
<p>Consider the sets
<span class="math-container">$$
\begin{align*}
&amp;UC(y) = \{x: x \succeq y\},\\
&amp;LC(y) = \{x: y \succeq x\},\\
&amp;SUC(y) = \{x: x \succ y\},\\
&amp;SLC(y) = \{x: y \succ x\}.
\end{align*}
$$</span>
We have the following equivalences:</p>
<ul>
<li>1 is equivalent to the assumption that <span class="math-container">$SUC(y)$</span> is open.</li>
<li>2 is equivalent to the assumption that <span class="math-container">$SLC(y)$</span> is open</li>
<li>3 is equivalent to the assumption that <span class="math-container">$UC(y)$</span> is closed</li>
<li>4 is equivalent to the assumption that <span class="math-container">$LC(y)$</span> is closed.</li>
</ul>
<p>We also have that:</p>
<ul>
<li><span class="math-container">$SUC(y)$</span> is the complement of <span class="math-container">$LC(y)$</span>.</li>
<li><span class="math-container">$SLC(y)$</span> is the complement of <span class="math-container">$UC(y)$</span>.</li>
</ul>
<p>We also know that a set is open if and only if its complement is closed. From this it follows that <span class="math-container">$1$</span> is equivalent to <span class="math-container">$4$</span> and <span class="math-container">$2$</span> is equivalent to <span class="math-container">$3$</span>.</p>
<p>Alternatively you can also proof it directly</p>
<p><strong>Th</strong> 1 is equivalent to 4</p>
<p><em>proof</em> (<span class="math-container">$1 \to 4$</span>) Let <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y \succeq x_n$</span>. Towards a contradiction assume that <span class="math-container">$y \not \succeq x$</span> which means that <span class="math-container">$x \succ y$</span>. Then we know that there exists an open ball <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> such that for all <span class="math-container">$z \in B_r$</span>, <span class="math-container">$z \succ y$</span>. But for <span class="math-container">$n$</span> large enough we have that <span class="math-container">$x_n \in B_r$</span>, so for <span class="math-container">$n$</span> large enough also <span class="math-container">$x_n \succ y$</span>, which is a contradiction.</p>
<p><span class="math-container">$(4 \to 1)$</span>. Assume that <span class="math-container">$x \succ y$</span> and assume that for all open balls <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> there exists an <span class="math-container">$z_r$</span> such that <span class="math-container">$z_r \not \succ y$</span>, which means that <span class="math-container">$y \succeq z_r$</span>. Take a sequence <span class="math-container">$r_n \to 0$</span> which generates a sequence <span class="math-container">$z_n$</span> with <span class="math-container">$y \succeq z_n$</span> and <span class="math-container">$z_n \to x$</span>. Assumption 4 then gives that <span class="math-container">$y \succeq x$</span>, which gives the desired contradiction.</p>
<p>The equivalence between 2 and 3 can be shown in a similar way.</p>
<p>... continue answer</p>
<h2>Other continuity conditions</h2>
<p>Consider in addition the following 2 conditions</p>
<ol start="5">
<li><p>If  <span class="math-container">$x \succ y$</span> there is a ball <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> and a ball <span class="math-container">$B_s$</span> around <span class="math-container">$y$</span> such that for all <span class="math-container">$z \in B_r$</span> and <span class="math-container">$w \in B_s$</span>, <span class="math-container">$z \succ w$</span>.</p>
</li>
<li><p>If <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y_n \to y$</span> and for all <span class="math-container">$n$</span>, <span class="math-container">$y_n \succeq x_n$</span> then <span class="math-container">$y \succeq x$</span>.</p>
</li>
</ol>
<p>Let us first show that these two are equivalent</p>
<p><strong>Th</strong> <span class="math-container">$5$</span> is equivalent to <span class="math-container">$6$</span>.</p>
<p><em>proof</em> Assume that <span class="math-container">$5$</span> holds. Let <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y_n \to y$</span> with <span class="math-container">$y_n \succeq x_n$</span> for all <span class="math-container">$n$</span>. Towards a contradiction assume that <span class="math-container">$y \not \succeq x$</span>. Then <span class="math-container">$x \succ y$</span>. As such, there should be a ball <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> and <span class="math-container">$B_s$</span> around <span class="math-container">$y$</span> such that for all <span class="math-container">$z \in B_r$</span> and <span class="math-container">$w \in B_s$</span>, <span class="math-container">$z \succ w$</span>. Now, for all <span class="math-container">$n$</span> big enough <span class="math-container">$x_n \in B_r$</span> and <span class="math-container">$y_n \in B_s$</span>. As such, for <span class="math-container">$n$</span> big enough <span class="math-container">$x_n \succ y_n$</span>, a contradiction.</p>
<p>For the reverse, assume that <span class="math-container">$6$</span> holds <span class="math-container">$x \succ y$</span> and, towards a contradiction for all balls <span class="math-container">$B_r$</span> around <span class="math-container">$x$</span> and <span class="math-container">$B_s$</span> around <span class="math-container">$y$</span>, there are <span class="math-container">$z \in B_r$</span> and <span class="math-container">$w \in B_s$</span> such that <span class="math-container">$w \succeq z$</span>. Let <span class="math-container">$r_n \to 0$</span> and <span class="math-container">$s_n \to 0$</span> then we can generate a sequence <span class="math-container">$z_n \in B_{r_n}$</span> and <span class="math-container">$w_n \in B_{s_n}$</span> such that <span class="math-container">$z_n \to x$</span>, <span class="math-container">$w_n \to y$</span>. And for all <span class="math-container">$n$</span>, <span class="math-container">$w_n \succeq z_n$</span>. this gives (by <span class="math-container">$6$</span>) that <span class="math-container">$y \succeq x$</span>, a contradiction.</p>
<h2>Connecting the various conditions together</h2>
<p>Remember from the first part the following two conditions:</p>
<ol start="7">
<li><p><span class="math-container">$\forall x, y$</span>, if <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y \succeq x_n$</span> for all <span class="math-container">$n$</span> then <span class="math-container">$y \succeq x$</span> <span class="math-container">$\leftrightarrow$</span> <span class="math-container">$\forall y$</span>, <span class="math-container">$LC(y)$</span> is closed, <span class="math-container">$\leftrightarrow$</span> <span class="math-container">$\forall y$</span>, <span class="math-container">$SUC(y)$</span> is open</p>
</li>
<li><p><span class="math-container">$\forall x, y$</span> if <span class="math-container">$y_n \to y$</span> and <span class="math-container">$y_n \succeq x$</span> for all <span class="math-container">$n$</span> then <span class="math-container">$y \succeq x$</span> <span class="math-container">$\leftrightarrow$</span> <span class="math-container">$\forall x$</span>, <span class="math-container">$UC(x)$</span> is closed <span class="math-container">$\leftrightarrow$</span> <span class="math-container">$\forall x$</span>, <span class="math-container">$SLC(x)$</span> is open</p>
</li>
</ol>
<p>The following result shows the equivalence between <span class="math-container">$5$</span> (or <span class="math-container">$6$</span>)<span class="math-container">$ and $</span>7<span class="math-container">$ and $</span>8$. It is also exercise 3.C.3 in MWG (if you would be interested). The proof is not so self-evident.</p>
<p><strong>Th</strong> <span class="math-container">$5$</span> (or equivalently <span class="math-container">$6$</span>) hold if and only if both <span class="math-container">$7$</span> and <span class="math-container">$8$</span> hold.</p>
<p><em>proof</em> That <span class="math-container">$5$</span> (or <span class="math-container">$6$</span>) imply <span class="math-container">$7$</span> and <span class="math-container">$8$</span> is obvious as we can take the constant sequence. For the reverse, assume that <span class="math-container">$7$</span> and <span class="math-container">$8$</span> hold and that <span class="math-container">$6$</span> is not true. This means that <span class="math-container">$x_n \to x$</span>, <span class="math-container">$y_n \to y$</span>, <span class="math-container">$y_n \succeq x_n$</span> for all <span class="math-container">$n$</span> and <span class="math-container">$x \succ y$</span>. Then as <span class="math-container">$SUC(y)$</span> is open, there is an <span class="math-container">$N_1$</span> such that for all <span class="math-container">$n \ge N_1$</span>:
<span class="math-container">$$
x_n \succ y.
$$</span>
As <span class="math-container">$SLC(x)$</span> is open, there is also an <span class="math-container">$N_2$</span> such that for all <span class="math-container">$n \ge N_2$</span>:
<span class="math-container">$$
x \succ y_n.
$$</span>
There are two possible cases.</p>
<ol>
<li><p>There is an <span class="math-container">$N_3$</span> such that for all <span class="math-container">$n \ge N_3$</span>:
<span class="math-container">$$
   x_n \succeq x.
   $$</span></p>
</li>
<li><p>There is a subsequence <span class="math-container">$k(n)$</span> such that for all <span class="math-container">$n$</span><br />
<span class="math-container">$$
   x \succ x_{k(n)}.
   $$</span></p>
</li>
</ol>
<p>If <span class="math-container">$1.$</span> is the case, then for all <span class="math-container">$n \ge \max\{N_3, N_2\}$</span>,
<span class="math-container">$$
x_n \succeq x \succ y_n,
$$</span>
a contradiction.</p>
<p>If <span class="math-container">$2.$</span> is the case, then we can find an <span class="math-container">$m$</span> such that <span class="math-container">$k(m) \ge N_1$</span>. Then:
<span class="math-container">$$
x \succ x_{k(m)} \succ y
$$</span>
As <span class="math-container">$SUC(x_{k(m)})$</span> is open and <span class="math-container">$x_n \to x$</span>, we know there is an <span class="math-container">$N_4$</span> such that for all <span class="math-container">$n \ge N_4$</span>:
<span class="math-container">$$
y_n \succeq x_n \succ x_{k(m)} \succ y
$$</span>
Taking the limit for <span class="math-container">$n \to \infty$</span> we see that:
<span class="math-container">$$
y \succeq x_{k(m)} \succ y.
$$</span>
again a contradiction.</p>

CC BY-SA 4.0
2021-06-24T21:26:28.207Z
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.)
not tdm's twin
CC BY-SA 4.0
2021-06-26T17:26:10.527Z
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 $y \prec s$.
not tdm's twin
CC BY-SA 4.0
2021-06-26T18:50:16.660Z
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.
not tdm's twin
CC BY-SA 4.0
2021-06-27T13:38:58.703Z
@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
CC BY-SA 4.0
2021-06-27T14:43:45.777Z
@dictatemetokcus btw, if you want me to delete the answer, just let me know (no problem).
tdm
CC BY-SA 4.0
2021-06-29T11:22:02.257Z
Right, the fact that you were not responding at all even to my previous questions and yet continuously answering me got me triggered. Not that I did not appreciate your efforts, I absolutely did. I have a more concise solution now, and I will add it as an answer later. Nevertheless, for your efforts, I believe I should pass the reputation to you.
not tdm's twin
CC BY-SA 4.0
2021-06-24T17:25:31.503Z
2021-06-27T13:35:31.673Z
2021-06-27T13:35:31.673Z
tdm
<p>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.</p>
<p>Def. 1: A preference relation  <span class="math-container">$(\succeq)$</span> on <span class="math-container">$X$</span> is continuous if for <span class="math-container">$r \succ s$</span>, there exists balls <span class="math-container">$B_r$</span> and <span class="math-container">$B_s$</span> around <span class="math-container">$r$</span> and <span class="math-container">$s$</span> (respectively) such that for every <span class="math-container">$x$</span> in <span class="math-container">$B_r$</span> and <span class="math-container">$y$</span> in <span class="math-container">$B_s$</span>, we have <span class="math-container">$x \succ y$</span>.</p>
<p>Def. 2: <span class="math-container">$\text{(Part 1)}$</span> If <span class="math-container">$x_n \rightarrow  x$</span> and <span class="math-container">$x_n \succeq y$</span> <span class="math-container">$\forall \ n$</span> then <span class="math-container">$x \succeq y$</span>, and <span class="math-container">$\text{(Part 2)}$</span> if <span class="math-container">$x_n \preceq y$</span> <span class="math-container">$\forall \ n$</span> then <span class="math-container">$x \preceq y$</span>.</p>
<hr />
<p>My questions:</p>
<ol>
<li><p>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?</p>
</li>
<li><p>Many authors simply write that <em>&quot;if A is more preferable to B and B is 'sufficiently close' to C, then A is preferable to C&quot;</em>. 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:
<span class="math-container">$$\text{(Part 1) } (A \succeq B) \land (A \rightarrow C) \implies C \succeq B \\ \text{(Part 2) } (A \succeq B) \land (B \rightarrow C) \implies A \succeq C$$</span></p>
</li>
</ol>

CC BY-SA 4.0
2021-06-24T14:26:42.390Z
[Rubinstein's lecture note](https://arielrubinstein.org/gt/arielDocs/downloadme.asp?id=5#page=30) actually have a proof of the equivalence of the two definitions.
Herr K.
<microeconomics>
Equivalence of the definitions of the Axiom of Continuity
not tdm's twin
# Equivalence of the definitions of the Axiom of Continuity

(Asked by: username_0 on 2021-06-24)

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

**Tags:** microeconomics

> **Comments:**
> **username_1**: [Rubinstein's lecture note](https://arielrubinstein.org/gt/arielDocs/downloadme.asp?id=5#page=30) actually have a proof of the equivalence of the two definitions.  

---

## 1 answers

### Answer by username_2 on 2021-06-24. Score: 2 | Accepted answer

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

5. 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 $w \in B\_s$, $z \succ w$.
6. 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:

7. $\forall x, y$, 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
8. $\forall x, y$ 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.

> **Comments:**
> **username_0**: 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.)  
> **username_0**: 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 $y \prec s$.  
> **username_0**: 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.  
> **username_2**: @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. (?)  
> **username_2**: @dictatemetokcus btw, if you want me to delete the answer, just let me know (no problem).  
> **username_0**: Right, the fact that you were not responding at all even to my previous questions and yet continuously answering me got me triggered. Not that I did not appreciate your efforts, I absolutely did. I have a more concise solution now, and I will add it as an answer later. Nevertheless, for your efforts, I believe I should pass the reputation to you.  

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