<p>Without loss of generality, suppose <span class="math-container">$\mathbf x=(x_1,0)$</span> where <span class="math-container">$x_1&gt;0$</span>. Consider the following sequence
<span class="math-container">\begin{equation}
\mathbf y^n=\left(x_1\Bigl(1-\frac1n\Bigr),\,\frac1n\right).
\end{equation}</span>
Clearly, <span class="math-container">$\mathbf y^n\gg\mathbf 0$</span> for all <span class="math-container">$n\in\mathbb N$</span>, and thus <span class="math-container">$\mathbf y^n\succsim\mathbf 0$</span> by monotonicity. It is also the case that <span class="math-container">$\lim_{n\to\infty}\mathbf y^n= \mathbf x$</span>. Hence by continuity, we have <span class="math-container">$\mathbf x\succsim\mathbf 0$</span>.</p>

CC BY-SA 4.0
2021-02-02T17:07:05.887Z
2021-02-02T17:07:05.887Z
Herr K.
<p>Suppose a consumer has a preference ordering <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> that is complete. Show that if preferences are continuous and monotone, then <span class="math-container">$x\succsim0$</span> for any <span class="math-container">$x\in\mathbb{R}_{+}^{N}$</span>, where <span class="math-container">$0$</span> is the <span class="math-container">$0$</span> vector in <span class="math-container">$R_{+}^{N}.$</span></p>
<p>My approach so far is the following.</p>
<p>Case 1. <span class="math-container">$x$</span> has more of all commodities, in which case the result follows from monotonicity.</p>
<p>Case 2. <span class="math-container">$x$</span> has more of only one commodity, in which case monotonicity is not sufficient. We want to show that <span class="math-container">$x$</span> is in the upper contour set of <span class="math-container">$0$</span>, i.e. that <span class="math-container">$x\in\{\phi\in X:\phi\succsim0\}$</span>. If we can show that there exists some sequence of bundles in this set with bundle <span class="math-container">$x$</span> as its limit, then the result would follow from continuity. But I am not sure how to do this.</p>
<p>Case 3. <span class="math-container">$x:=0$</span>. Trivial.</p>

<microeconomics><general-equilibrium>
Monotonicity and continuity implies that all bundles are weakly preferred to 0
Charles
# Monotonicity and continuity implies that all bundles are weakly preferred to 0

(Asked by: username_0 on 2021-02-02)

Suppose a consumer has a preference ordering $\succsim$ on $X$ that is complete. Show that if preferences are continuous and monotone, then $x\succsim0$ for any $x\in\mathbb{R}\_{+}^{N}$, where $0$ is the $0$ vector in $R\_{+}^{N}.$

My approach so far is the following.

Case 1. $x$ has more of all commodities, in which case the result follows from monotonicity.

Case 2. $x$ has more of only one commodity, in which case monotonicity is not sufficient. We want to show that $x$ is in the upper contour set of $0$, i.e. that $x\in\{\phi\in X:\phi\succsim0\}$. If we can show that there exists some sequence of bundles in this set with bundle $x$ as its limit, then the result would follow from continuity. But I am not sure how to do this.

Case 3. $x:=0$. Trivial.

**Tags:** microeconomics, general-equilibrium

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## 1 answers

### Answer by username_1 on 2021-02-02. Score: 3

Without loss of generality, suppose $\mathbf x=(x\_1,0)$ where $x\_1>0$. Consider the following sequence
\begin{equation}
\mathbf y^n=\left(x\_1\Bigl(1-\frac1n\Bigr),\,\frac1n\right).
\end{equation}
Clearly, $\mathbf y^n\gg\mathbf 0$ for all $n\in\mathbb N$, and thus $\mathbf y^n\succsim\mathbf 0$ by monotonicity. It is also the case that $\lim\_{n\to\infty}\mathbf y^n= \mathbf x$. Hence by continuity, we have $\mathbf x\succsim\mathbf 0$.


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