$x<\sup Y$ if and only if $\exists y\in Y: x<y$

Question

Let $(X,<)$ be a nonempty partially ordered set, $Y$ be a nonempty subset of $X$, and $x\in X$. Then $x<\sup Y \iff \exists y\in Y: x<y$.

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My attempt:

1. $x<\sup Y \implies \exists y\in Y: x<y$

If not, $\forall y\in Y: y\le x$. Then $\sup X\le x$. This is a contradiction. Thus $x<\sup Y \implies \exists y\in Y: x<y$.

2. $\exists y\in Y: x<y \implies x<\sup Y$

We have $y\in Y \implies y \le \sup Y$. Moreover, $x<y$. Then $x<y\le \sup Y$. Hence $x<\sup Y$.

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Please verify if my attempt contains logical gaps or flaws! Thank your for your help!

Answer 1

From left to right is false in general for partial orders. Consider $\mathbb{N}^2$ with the product order: $\langle k,l\rangle\preceq\langle m,n\rangle$ iff $k\le m$ and $l\le n$. Then $\langle 2,2\rangle\prec\langle 3,3\rangle$ and $\langle3,3\rangle=\sup\{\langle1,3\rangle,\langle3,1\rangle\}$ but the conclusion does not follow.

From right to left is fine.