# 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 $$x_1>0$$. Consider the following sequence $$$$\mathbf y^n=\left(x_1\Bigl(1-\frac1n\Bigr),\,\frac1n\right).$$$$ Clearly, $$\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$$.