Monotonicity and continuity implies that all bundles are weakly preferred to 0 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.
Without loss of generality, suppose $\mathbf x=(x_1,0)$ where $x_1&gt;0$. 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 $\mathbf x\succsim\mathbf 0$.