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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?

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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, $x\succeq y$.

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