# Equivalence of two definitions of monotone preference

In MWG, the definition of weak preference is for all $$x,y \in X$$, $$y>>x$$ implies $$y\succ x$$ . But I have read some other articles that define weak preference as $$y\geq x\implies y\succeq x$$. Notice here, $$y\geq x$$ means $$y_i\geq x_i$$ for all $$i$$, and $$y>>x$$ means $$y_i>x_i$$ for all $$i$$.

These two definitions are not directly equivalent. I think the second one should imply the first MWG one, but I don't know how to prove it. It is clear that if $$y>>x$$ then $$y\geq x$$ so $$y\succeq x$$. But then I need to prove $$\lnot(x\succeq y)$$, I don't know how to approach. Additionally, from Understanding the definition of monotone, one of the answers gives a proof that if the preference is continuous and rational, then $$y>>x\implies y\succ x$$ implies $$y\geq x\implies y\succeq x$$. But I don't quite understand the proof. How should I approach this question if I want to use continuous definition of closed lower and upper contour sets?

Let $$\succeq$$ be a relation on $$\mathbb{R}^l_+$$ such that $$x\gg y$$ implies $$x\succeq y$$ for all $$x,y\in\mathbb{R}^l_+$$, and such that all upper contour sets are closed. Then $$x\geq y$$ implies $$x\succeq y$$.
Proof: Assume that $$x\geq y$$. Let $$\mathbf{1}=(1,1,\ldots,1)\in\mathbb{R}^l_+$$. Then for all $$n\geq 1$$, we have $$n^{-1}\mathbf{1}+x\gg y$$. By assumption, this implies $$n^{-1}\mathbf{1}+x\succ y$$ and, therefore, also $$n^{-1}\mathbf{1}+x\succeq y$$. Therefore, $$n^{-1}\mathbf{1}+x$$ lies in the upper contour set $$\{z\in \mathbb{R}^l_+\mid z\succeq y\}$$ for all $$n$$. But $$\lim_{n\to\infty}n^{-1}\mathbf{1}+x=x$$. Since upper contour sets are closed, the limit $$x$$ must also lie in $$\{z\in \mathbb{R}^l_+\mid z\succeq y\}.$$ Consequently, $$x\succeq y$$.