<p>We want to show that for $\succcurlyeq$ on $X$, <strong>Def 1 $\iff$ Def 2</strong></p>

<p>$\boxed \Longrightarrow$</p>

<p>Assume that $\succcurlyeq$ is continuous by <strong>Def 1</strong>.</p>

<p>Let us say $x \succ y$. Denote our open-balls as $B(x, r)$, an open ball around $x$ of radius $r$. Suppose $\forall n, \ \exists \ x^n \in B(x, \frac{1}{n}), \ y^n \in B(y, \frac{1}{n})$ such that $y^n \succcurlyeq x^n$. But then we have constructed $\{x^n\} \rightarrow x$ and $\{y^n\} \rightarrow y$, and by <strong>Def 1</strong>, $y \succcurlyeq x$, which is a contradiction.</p>

<p>$\boxed \Longleftarrow$</p>

<p>Assume that $\succcurlyeq$ is continuous by <strong>Def 2</strong>.</p>

<p>Let us take sequences ${x^n} \subset X$ and ${y^n} \subset X$ where $\forall n, \quad x^n \succcurlyeq y^n$ and $\lim_{n \rightarrow \infty} x^n = x, \quad \lim_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$ for $n \in \mathbb{N}$,</p>

<p>BUT $x \succcurlyeq y$ is false instead of true. We want to show this leads to a contradiction.</p>

<p>If $x \succcurlyeq y$ is false, (so $y \succ x$) then $\exists B_x, B_y$ such that $\forall y' \in B_y, x' \in B_x$, we have $y' \succ x'$. Because $\{x^n\} \rightarrow x, \{y^n\} \rightarrow y$, there exists $N$ large enough such that $\forall n&gt;N$, we have $y^n \in B_y, x^n \in B_x$.</p>

<p>Thus $\forall n &gt; N$, we have $y^n \succ x^n$, which contradicts $\forall n, \ x^n \succcurlyeq y^n$.</p>

CC BY-SA 3.0
2016-09-14T21:23:35.047Z
2016-09-14T21:23:35.047Z
Kitsune Cavalry
<p>We have two definitions of the continuity of preferences:</p>
<blockquote>
<p><strong>Def 1:</strong> <span class="math-container">$\succcurlyeq$</span> is continuous if for any sequences <span class="math-container">$\{x^n\} \subset X$</span> and <span class="math-container">$\{y^n\} \subset X$</span>, then <span class="math-container">$n \in \mathbb{N}$</span> such that,</p>
<ul>
<li><span class="math-container">$\forall n, \quad x^n \succcurlyeq y^n$</span></li>
<li><span class="math-container">$\lim_{n \rightarrow \infty} x^n = x, \quad \lim_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$</span></li>
</ul>
<p>then <span class="math-container">$x \succcurlyeq y$</span></p>
</blockquote>
<p>and</p>
<blockquote>
<p><strong>Def 2:</strong> <span class="math-container">$\succcurlyeq$</span> is continuous if whenever <span class="math-container">$x \succ y$</span>, <span class="math-container">$\exists \ B_x, \ B_y$</span>, open balls around <span class="math-container">$x, y$</span>, such that <span class="math-container">$\forall \ x' \in B_x, \ y' \in B_y$</span>, then we have <span class="math-container">$x' \succ y'$</span>.</p>
</blockquote>
<p>Show that the following definitions are equivalent to each other.</p>

<microeconomics><preferences><proof>
Equivalence of Definitions of Continuity of Preferences
Kitsune Cavalry
# Equivalence of Definitions of Continuity of Preferences

(Asked by: username_0 on 2016-09-14)

We have two definitions of the continuity of preferences:

> **Def 1:** $\succcurlyeq$ is continuous if for any sequences $\{x^n\} \subset X$ and $\{y^n\} \subset X$, then $n \in \mathbb{N}$ such that,
>
> * $\forall n, \quad x^n \succcurlyeq y^n$
> * $\lim\_{n \rightarrow \infty} x^n = x, \quad \lim\_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$
>
> then $x \succcurlyeq y$

and

> **Def 2:** $\succcurlyeq$ is continuous if whenever $x \succ y$, $\exists \ B\_x, \ B\_y$, open balls around $x, y$, such that $\forall \ x' \in B\_x, \ y' \in B\_y$, then we have $x' \succ y'$.

Show that the following definitions are equivalent to each other.

**Tags:** microeconomics, preferences, proof


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## 1 answers

### Answer by username_0 on 2016-09-14. Score: 3 | Accepted answer

We want to show that for $\succcurlyeq$ on $X$, **Def 1 $\iff$ Def 2**

$\boxed \Longrightarrow$

Assume that $\succcurlyeq$ is continuous by **Def 1**.

Let us say $x \succ y$. Denote our open-balls as $B(x, r)$, an open ball around $x$ of radius $r$. Suppose $\forall n, \ \exists \ x^n \in B(x, \frac{1}{n}), \ y^n \in B(y, \frac{1}{n})$ such that $y^n \succcurlyeq x^n$. But then we have constructed $\{x^n\} \rightarrow x$ and $\{y^n\} \rightarrow y$, and by **Def 1**, $y \succcurlyeq x$, which is a contradiction.

$\boxed \Longleftarrow$

Assume that $\succcurlyeq$ is continuous by **Def 2**.

Let us take sequences ${x^n} \subset X$ and ${y^n} \subset X$ where $\forall n, \quad x^n \succcurlyeq y^n$ and $\lim\_{n \rightarrow \infty} x^n = x, \quad \lim\_{n \rightarrow \infty} y^n = y \quad (\text{where $x, y \in X)$}$ for $n \in \mathbb{N}$,

BUT $x \succcurlyeq y$ is false instead of true. We want to show this leads to a contradiction.

If $x \succcurlyeq y$ is false, (so $y \succ x$) then $\exists B\_x, B\_y$ such that $\forall y' \in B\_y, x' \in B\_x$, we have $y' \succ x'$. Because $\{x^n\} \rightarrow x, \{y^n\} \rightarrow y$, there exists $N$ large enough such that $\forall n>N$, we have $y^n \in B\_y, x^n \in B\_x$.

Thus $\forall n > N$, we have $y^n \succ x^n$, which contradicts $\forall n, \ x^n \succcurlyeq y^n$.


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