<p>Usually the continuity axiom has two parts</p>
<ul>
<li>if <span class="math-container">$x_n \to x$</span> and <span class="math-container">$x_n \succeq y$</span> for all <span class="math-container">$n$</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">$x_n \preceq y$</span> for all <span class="math-container">$n$</span> then <span class="math-container">$x \preceq y$</span>.</li>
</ul>
<blockquote>
<p>I don't get the idea: how is this equivalent to <span class="math-container">$(A \succeq B) \wedge (A$</span> sufficently close to <span class="math-container">$C$</span>) <span class="math-container">$\to $</span> <span class="math-container">$C \succeq B$</span>.</p>
</blockquote>
<p>It is not. Take for example the case where preferences are strictly montonic which means that if <span class="math-container">$x &gt; y$</span> then <span class="math-container">$x \succ y$</span> (more is always better). Then if we take two indifferent bundles <span class="math-container">$x \sim y$</span> we have that for any <span class="math-container">$z &lt; x$</span> no matter how close to <span class="math-container">$x$</span> that <span class="math-container">$z \prec y$</span>, as <span class="math-container">$z \prec x \sim y$</span>.</p>
<p>What is true under continuity, however, is the following:</p>
<p><strong>Th</strong> if <span class="math-container">$x \succ y$</span> then there exists an <span class="math-container">$\epsilon &gt; 0$</span> such that for all <span class="math-container">$z$</span> that are <span class="math-container">$\epsilon$</span>-close to <span class="math-container">$x$</span>, we have <span class="math-container">$z \succ y$</span>.</p>
<blockquote>
<p><em>proof</em> The proof is by contradiction. Assume not, then for all <span class="math-container">$\varepsilon &gt; 0$</span> there is a <span class="math-container">$z$</span> which is <span class="math-container">$\epsilon$</span>-close to <span class="math-container">$x$</span> and <span class="math-container">$z \succeq y$</span>. Take a decreasing sequence of such <span class="math-container">$\epsilon = 1, 1/2, 1/3,\ldots, 1/n, \ldots$</span>. Then for all <span class="math-container">$n$</span> there is a <span class="math-container">$z_n$</span> which is <span class="math-container">$1/n$</span>-close to <span class="math-container">$x$</span> and <span class="math-container">$z_n \succeq y$</span>. We have that <span class="math-container">$z_n \to x$</span>, so by continuity <span class="math-container">$x \succeq y$</span>, a contradiction.</p>
</blockquote>
<blockquote>
<p>Similarly, what if only <span class="math-container">$x_{odd} \succeq y$</span> (and <span class="math-container">$x_{even} \preceq y$</span>) and <span class="math-container">$(x_i)_{i \ge 1} \to x$</span> holds, can't then <span class="math-container">$x \succeq y$</span>?</p>
</blockquote>
<p>Yes, this will indeed be the case.</p>
<p>To see this, assume that the condition hold. Take the subsequence of all odd terms <span class="math-container">$x_1, x_3, x_5,\ldots$</span> We have that for each <span class="math-container">$x_i$</span> in this term <span class="math-container">$x_i \succeq y$</span>. Also, along this sequence <span class="math-container">$x_i \to x$</span>, so by continuity <span class="math-container">$x \succeq y$</span>.</p>
<p>The case is even stronger as we can also consider the subsequence of even terms <span class="math-container">$x_2, x_4, \ldots$</span>. For each <span class="math-container">$x_i$</span> in this sequence <span class="math-container">$x_i \preceq y$</span> and <span class="math-container">$x_i \to x$</span>. As such also <span class="math-container">$ x \preceq y$</span>. This means that both <span class="math-container">$x \succeq y$</span> and <span class="math-container">$x \preceq y$</span> giving that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are indifferent: <span class="math-container">$x \sim y$</span>.</p>
<blockquote>
<p>In the above example, the axiom says it can be true although it's not necessary for <span class="math-container">$x \succeq y$</span> to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?</p>
</blockquote>
<p>I don't really understand this question.</p>
<p>Fix an <span class="math-container">$y$</span>. Continuity then basically requires that there is not a sudden jump from the region where you are strictly better of than <span class="math-container">$y$</span>, i.e. where <span class="math-container">$x \succ y$</span>,  to the region where you are strictly worse of than with <span class="math-container">$y$</span>,i.e. where <span class="math-container">$x \prec y$</span>.</p>
<p>In other words if you move in a continuous way from an element <span class="math-container">$x$</span> with <span class="math-container">$x \succ y$</span> to an element <span class="math-container">$x'$</span> with <span class="math-container">$x' \prec y$</span> then at some point you will find an element <span class="math-container">$x''$</span> on this path with <span class="math-container">$x'' \sim y$</span>.</p>
<p>An alternative intuition is to look at the upper contour sets:
<span class="math-container">$$
U(y) = \{x: x \succeq y\}
$$</span>
and the lower contour sets
<span class="math-container">$$
L(y) = \{x: x \preceq y\}.
$$</span>
Continuity requires that both these sets are closed.</p>

CC BY-SA 4.0
2021-06-23T15:07:14.877Z
Hi, thanks for your solution. Could you tell me why some articles define the continuity axiom as "A more preferable to to B and B sufficiently close to C implies A more preferable to C" and some articles state in the way you have. This is in regard to your answer to my first question.
not tdm's twin
CC BY-SA 4.0
2021-06-23T15:16:28.230Z
Wait, I reversed the statement there completely. It's only defined as in the previous comment. I thought it holds both ways. So it's ONLY $(A \succeq B) \land (B \to C) \implies A \succeq C$.
not tdm's twin
CC BY-SA 4.0
2021-06-23T15:20:29.570Z
Could you tell me if that is correct?
not tdm's twin
CC BY-SA 4.0
2021-06-23T15:21:46.447Z
@dictatemetokcus That's the second part of the definition of continuity. See the top of my answer.
tdm
CC BY-SA 4.0
2021-06-23T15:34:11.320Z
Do you mean $(A \succeq B) \land (A \to C) \implies C \succeq B$ is same as your first definition and $(A \succeq B) \land (B \to C) \implies A \succeq C$ as your second definition?
not tdm's twin
CC BY-SA 4.0
2021-06-23T16:08:55.170Z
What I don't understand now is: if we break the definition into two parts, namely (1) A close to C and A $\succeq$ B then : C $\succeq$ B and (2) B close to C and A $\succeq$ B then : A $\succeq$ C, then how is your example equivalent to (2)? The "~" can be treated as "$\succeq$" (it does hold both ways).
not tdm's twin
CC BY-SA 4.0
2021-06-23T16:10:06.960Z
I am sorry I don't think I understand your solution well yet. I would really appreciate if you could elaborate it further along the lines I wrote in the comments above.
not tdm's twin
CC BY-SA 4.0
2021-06-23T14:10:35.807Z
2021-06-23T14:10:35.807Z
tdm
<p>Let <span class="math-container">$x^{1}, x^{2}, \cdots \to x$</span> where each <span class="math-container">$x^{i}$</span> and <span class="math-container">$x$</span> are elements of the set of consumption bundle or the choice set <span class="math-container">$X$</span>. If <span class="math-container">$x^{i} \succeq y$</span> for each <span class="math-container">$i \geq 1$</span> then <span class="math-container">$x \succeq y$</span>. This is the continuity axiom of preference.</p>
<p>Since this is an axiom, I don't think there is a proof.</p>
<p><strong>Q1.</strong> I don't get the idea: how is this equivalent to <span class="math-container">$(A \succeq B) \ \land \ (A \text{ sufficiently close to } C) \implies C \succeq B$</span> ?</p>
<p>Similarly, what if only <span class="math-container">$x^{\text{odd}} \succeq y$</span> (and <span class="math-container">$x^\text{even} \preceq y$</span>) and <span class="math-container">$(x^{i})_{i \geq 1} \to x$</span> holds, can't then <span class="math-container">$x \succeq y$</span>?</p>
<p><strong>Q2.</strong> In the above example, the axiom says it can be true although it's not <em>necessary</em> for <span class="math-container">$x \succeq y$</span> to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?</p>
<hr />
<p>Edit: I am still not very clear with this; hence the answer below has not been accepted. I have added a few comments below the answer.</p>

CC BY-SA 4.0
2021-06-23T17:25:54.753Z
Continuity is a preservation property. Suppose we have a mapping between two sets. Continuity ensures that a distance/magnitude/ordering in one set; is also sufficiently close in another set.
EB3112
CC BY-SA 4.0
2021-06-23T18:34:57.537Z
@EB3112 I understand that; I am confused with the technicalities. I have written them below the answer, waiting for someone to explain that part to me.
not tdm's twin
<microeconomics><preferences>
Understanding the Continuity axiom of preference
not tdm's twin
# Understanding the Continuity axiom of preference

(Asked by: username_0 on 2021-06-23)

Let $x^{1}, x^{2}, \cdots \to x$ where each $x^{i}$ and $x$ are elements of the set of consumption bundle or the choice set $X$. If $x^{i} \succeq y$ for each $i \geq 1$ then $x \succeq y$. This is the continuity axiom of preference.

Since this is an axiom, I don't think there is a proof.

**Q1.** I don't get the idea: how is this equivalent to $(A \succeq B) \ \land \ (A \text{ sufficiently close to } C) \implies C \succeq B$ ?

Similarly, what if only $x^{\text{odd}} \succeq y$ (and $x^\text{even} \preceq y$) and $(x^{i})\_{i \geq 1} \to x$ holds, can't then $x \succeq y$?

**Q2.** In the above example, the axiom says it can be true although it's not *necessary* for $x \succeq y$ to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?

---

Edit: I am still not very clear with this; hence the answer below has not been accepted. I have added a few comments below the answer.

**Tags:** microeconomics, preferences

> **Comments:**
> **username_1**: Continuity is a preservation property. Suppose we have a mapping between two sets. Continuity ensures that a distance/magnitude/ordering in one set; is also sufficiently close in another set.  
> **username_0**: @username_1 I understand that; I am confused with the technicalities. I have written them below the answer, waiting for someone to explain that part to me.  

---

## 1 answers

### Answer by username_2 on 2021-06-23. Score: 4

Usually the continuity axiom has two parts

* if $x\_n \to x$ and $x\_n \succeq y$ for all $n$ then $x \succeq y$.
* if $x\_n \to x$ and $x\_n \preceq y$ for all $n$ then $x \preceq y$.

> I don't get the idea: how is this equivalent to $(A \succeq B) \wedge (A$ sufficently close to $C$) $\to $ $C \succeq B$.

It is not. Take for example the case where preferences are strictly montonic which means that if $x > y$ then $x \succ y$ (more is always better). Then if we take two indifferent bundles $x \sim y$ we have that for any $z < x$ no matter how close to $x$ that $z \prec y$, as $z \prec x \sim y$.

What is true under continuity, however, is the following:

**Th** if $x \succ y$ then there exists an $\epsilon > 0$ such that for all $z$ that are $\epsilon$-close to $x$, we have $z \succ y$.

> *proof* The proof is by contradiction. Assume not, then for all $\varepsilon > 0$ there is a $z$ which is $\epsilon$-close to $x$ and $z \succeq y$. Take a decreasing sequence of such $\epsilon = 1, 1/2, 1/3,\ldots, 1/n, \ldots$. Then for all $n$ there is a $z\_n$ which is $1/n$-close to $x$ and $z\_n \succeq y$. We have that $z\_n \to x$, so by continuity $x \succeq y$, a contradiction.

> Similarly, what if only $x\_{odd} \succeq y$ (and $x\_{even} \preceq y$) and $(x\_i)\_{i \ge 1} \to x$ holds, can't then $x \succeq y$?

Yes, this will indeed be the case.

To see this, assume that the condition hold. Take the subsequence of all odd terms $x\_1, x\_3, x\_5,\ldots$ We have that for each $x\_i$ in this term $x\_i \succeq y$. Also, along this sequence $x\_i \to x$, so by continuity $x \succeq y$.

The case is even stronger as we can also consider the subsequence of even terms $x\_2, x\_4, \ldots$. For each $x\_i$ in this sequence $x\_i \preceq y$ and $x\_i \to x$. As such also $ x \preceq y$. This means that both $x \succeq y$ and $x \preceq y$ giving that $x$ and $y$ are indifferent: $x \sim y$.

> In the above example, the axiom says it can be true although it's not necessary for $x \succeq y$ to hold true. In this regard, can I get an example to understand why the (original) axiom makes sense in the real world (or at all, in general) and this does not?

I don't really understand this question.

Fix an $y$. Continuity then basically requires that there is not a sudden jump from the region where you are strictly better of than $y$, i.e. where $x \succ y$, to the region where you are strictly worse of than with $y$,i.e. where $x \prec y$.

In other words if you move in a continuous way from an element $x$ with $x \succ y$ to an element $x'$ with $x' \prec y$ then at some point you will find an element $x''$ on this path with $x'' \sim y$.

An alternative intuition is to look at the upper contour sets:
$$
U(y) = \{x: x \succeq y\}
$$
and the lower contour sets
$$
L(y) = \{x: x \preceq y\}.
$$
Continuity requires that both these sets are closed.

> **Comments:**
> **username_0**: Hi, thanks for your solution. Could you tell me why some articles define the continuity axiom as "A more preferable to to B and B sufficiently close to C implies A more preferable to C" and some articles state in the way you have. This is in regard to your answer to my first question.  
> **username_0**: Wait, I reversed the statement there completely. It's only defined as in the previous comment. I thought it holds both ways. So it's ONLY $(A \succeq B) \land (B \to C) \implies A \succeq C$.  
> **username_0**: Could you tell me if that is correct?  
> **username_2**: @dictatemetokcus That's the second part of the definition of continuity. See the top of my answer.  
> **username_0**: Do you mean $(A \succeq B) \land (A \to C) \implies C \succeq B$ is same as your first definition and $(A \succeq B) \land (B \to C) \implies A \succeq C$ as your second definition?  
> **username_0**: What I don't understand now is: if we break the definition into two parts, namely (1) A close to C and A $\succeq$ B then : C $\succeq$ B and (2) B close to C and A $\succeq$ B then : A $\succeq$ C, then how is your example equivalent to (2)? The "~" can be treated as "$\succeq$" (it does hold both ways).  
> **username_0**: I am sorry I don't think I understand your solution well yet. I would really appreciate if you could elaborate it further along the lines I wrote in the comments above.  

---

