Monotonicity and continuity implies that all bundles are weakly preferred to 0
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 $\mathbf x\succsim\mathbf 0$.