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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}\...


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