I am learning continuous poset by myself. I have conclusion as follows: If $P$ is a continuous poset w.r.t. Scott topology then there is $x\in P$ s.t. for any $y\in P$ and for any open sets $U_x$ and $U_y$ ($x\in U_x$ and $y\in U_y$), $y\in U_x$ or $x\in U_y$.
For instance $P$ is set of all real numbers. Then for all $x\in P$,$0\in U_x$ or $x\in U_0$.
If my conclusion is correct,please somebody help me to prove it. If it is false,please give me a counterexample
It’s not true in general.
Let $P=\Bbb R\times\Bbb R$, and set $\langle x_0,y_0\rangle\preceq\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. This is a continuous poset: the elements way below $\langle x_0,y_0\rangle$ are the $\langle x,y\rangle$ with $x<x_0$ and $y<y_0$, and they are a directed set with supremum $\langle x_0,y_0\rangle$.
Now fix $p=\langle x_0,y_0\rangle\in P$. Let $q=\langle x_0+2,y_0-2\rangle$. Let
$$U_p=\{\langle x,y\rangle:x>x_0-1\text{ and }y>y_0-1\}$$
and
$$U_q=\{\langle x,y\rangle:x>x_0+1\text{ and }y>y_0-3\}\;.$$
Then $U_p$ and $U_q$ are Scott open, $p\in U_p\setminus U_q$, and $q\in U_q\setminus U_p$. Thus, there is no $p\in P$ that acts like $0$ in your example.