Weakly Meet $s_{Z}$-continuity and $\delta_{Z}$-continuity

$(1)\Rightarrow(2)$ Assume that $D\in Z(P)$, and $D^{\delta}\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$. Then there exists an $m\in D^{\delta}$ with $m\in U$ and $m\leq x$. Since $P$ is weakly meet $s_{Z}$-continuous, we have $m\in cl_{\sigma^{Z}(P)}(\mathord{\downarrow}m\cap\mathord{\downarrow}D)$, which implies that $\mathord{\downarrow}D\cap\mathord{\downarrow}m\cap U\neq\emptyset$. Thus $\mathord{\downarrow}D\cap\mathord{\downarrow}x\cap U\neq\emptyset$ by $m\leq x$. So $D\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$ and $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$ holds.

$(2)\Rightarrow(1)$ For any $x\in P$, $D\in Z(P)$, if $x\in D^{\delta}$ and there is a $U\in\sigma^{Z}(P)$ such that $x\in U$, then by (2), $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$. Since $x\in D^{\delta}\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$, we have $D\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$, this means $\mathord{\downarrow}x\cap U\cap\mathord{\downarrow}D\neq\emptyset$. So $x\in cl_{\sigma^{Z}(P)}(\mathord{\downarrow}x\cap\mathord{\downarrow}D)$.

$(1)\Rightarrow(2)$ We first claim that $y=\vee(\mathord{\downarrow}y\cap\mathord{\downarrow}D)$ if $y\in D^{\delta}$. It is obvious that $y$ is an upper bound of $\mathord{\downarrow}y\cap\mathord{\downarrow}D$. Suppose $z$ is also an upper bound of $\mathord{\downarrow}y\cap\mathord{\downarrow}D$ and $y\nleqslant z$, that is, $y\in P\setminus\mathord{\downarrow}z$. Since $y\in cl_{\sigma^{Z}(P)}(\mathord{\downarrow}y\cap\mathord{\downarrow}D)$ by (1) and $P\setminus\mathord{\downarrow}z\in\sigma^{Z}(P)$, we have $(P\setminus\mathord{\downarrow}z)\cap\mathord{\downarrow}y\cap\mathord{\downarrow}D\neq\emptyset$. But this contracts the fact that $\mathord{\downarrow}y\cap\mathord{\downarrow}D\subseteq\mathord{\downarrow}z$. Thus $y=\vee(\mathord{\downarrow}y\cap\mathord{\downarrow}D)$. Now let $y_{0}=x\wedge\vee D$, then $y_{0}\in D^{\delta}$, which implies $y_{0}=\vee(\mathord{\downarrow}y_{0}\cap\mathord{\downarrow}D)$. Since $\mathord{\downarrow}y_{0}\cap\mathord{\downarrow}D=\mathord{\downarrow}\{x\wedge d:d\in D\}$, we have $y_{0}=\vee\{x\wedge d:d\in D\}$, that is, $x\wedge\vee D=\vee\{x\wedge d:d\in D\}$.

$(2)\Rightarrow(1)$ For any $x\in P$, $U\in\sigma^{Z}(P)$, we need to prove $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$. Assume $D\in Z(P)$ with $D^{\delta}\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$. Then there exists an $m\in U$, $m\leq x$ and $m\in D^{\delta}$. Thus $m\leq\vee D$ and $m=m\wedge\vee D=\vee\{m\wedge d:d\in D\}\in U$ by (2). Now for $m$, we define a monotone mapping $\varphi:P\rightarrow P$ by $\varphi(p)=m\wedge p$. Then $\varphi(D)=\{m\wedge d:d\in D\}\in Z(P)$. Hence, $m\wedge d_{0}\in U$ for some $d_{0}\in D$ as $U\in\sigma^{Z}(P)$, which implies that $D\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$, that is, $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$. So $P$ is weakly meet $s_{Z}$-continuous by Lemma 3.2.

$(1)\Rightarrow(2)$ By Lemma 3.4, we only need to prove that for any $A\in\Gamma^{Z}(P),\mathcal{D}\in Z(\Gamma^{Z}(P))$, $A\wedge(\vee\mathcal{D})=\vee\{A\wedge D:D\in\mathcal{D}\}$, that is, $A\cap cl_{\sigma^{Z}(P)}(\bigcup\mathcal{D})=cl_{\sigma^{Z}(P)}(\bigcup\{A\cap D:D\in\mathcal{D}\})$. Assume $x\in A\cap cl_{\sigma^{Z}(P)}(\bigcup\mathcal{D})$ and $U\in\sigma^{Z}(P)$ with $x\in U$. Then we have $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$ by Lemma 3.2. As $x\in cl_{\sigma^{Z}(P)}(\bigcup\mathcal{D})$, $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\cap D_{0}\neq\emptyset$ for some $D_{0}\in\mathcal{D}$, this means $A\cap U\cap D_{0}\neq\emptyset$ since $x\in A$ and $A$ is a lower set. Moreover, $(\bigcup\{A\cap D:D\in\mathcal{D}\})\cap U\neq\emptyset$. So $x\in cl_{\sigma^{Z}(P)}(\cup\{A\cap D:D\in\mathcal{D}\})$, and $A\cap cl_{\sigma^{Z}(P)}(\bigcup\mathcal{D})\subseteq cl_{\sigma^{Z}(P)}(\bigcup\{A\cap D:D\in\mathcal{D}\})$ holds. Obviously, the conversely inclusion holds.

$(2)\Rightarrow(1)$ It is sufficient to prove that $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$ for any $x\in P$, $U\in\sigma^{Z}(P)$. Let $D\in Z(P)$ with $D^{\delta}\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$. Then there exists an $m\in\mathord{\downarrow}x\cap U$ such that $m\in D^{\delta}$, which implies $\mathord{\downarrow}m\in\{\mathord{\downarrow}d:d\in D\}^{\delta}$. In addition, we know $\{\mathord{\downarrow}d:d\in D\}\in Z(\Gamma^{Z}(P))$ since the mapping $\psi:P\rightarrow\Gamma^{Z}(P)$ defined by $\psi(p)=\mathord{\downarrow}p$ is monotone. As $\Gamma^{Z}(P)$ is weakly meet $s_{Z}$-continuous, we have $\mathord{\downarrow}m\in cl_{\sigma^{Z}(\Gamma^{Z}(P))}(\mathord{\downarrow}\{\mathord{\downarrow}m\}\cap\mathord{\downarrow}\{\mathord{\downarrow}d:d\in D\})$. It is easy to verify that $\lozenge U=\{A\in\Gamma^{Z}(P):A\cap U\neq\emptyset\}\in\sigma^{Z}(\Gamma^{Z}(P))$ and $\mathord{\downarrow}m\in\lozenge U$. So $\lozenge U\cap\mathord{\downarrow}\{\mathord{\downarrow}m\}\cap\mathord{\downarrow}\{\mathord{\downarrow}d:d\in D\}\neq\emptyset$, that is, $C\in\Gamma^{Z}(P)$ belongs to this intersection. Moreover, there exists an element $c\in C\cap U$ satisfying $c\leq m\leq x$ and $c\leq d_{0}$ for some $d_{0}\in D$, this means $D\cap\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\neq\emptyset$. Hence, $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$, and $P$ is weakly meet $s_{Z}$-continuous.

Suppose $F=\{x_{1},x_{2},...,x_{n}\}$ and there exists an element $a\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)$, but $a\notin\cup\{\rotatebox{-90.0}{$\twoheadleftarrow$}x_{i}:i=1,2,...,n\}$. Then $x_{i}\leavevmode\hbox to16.67pt{\vbox to12.45pt{\pgfpicture\makeatletter\hbox{\hskip 3.333pt\lower-3.22398pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}{}}{{}}{} {}{}{}\pgfsys@moveto{0.0pt}{-2.0pt}\pgfsys@lineto{10.00002pt}{8.00002pt}\pgfsys@stroke\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{0.00002pt}{0.5pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$\ll$}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}_{Z}a$ for any $x_{i}\in F$, that is, there exists $D_{i}\in Z(P)$ such that $a\in D_{i}^{\delta}$, but $x_{i}\notin\mathord{\downarrow}D_{i}$, for $i=1,2,...,n$. For $D_{1}\in Z(P)$ with $a\in D_{1}^{\delta}$, $a\in cl_{\sigma^{Z}(P)}(\mathord{\downarrow}a\cap\mathord{\downarrow}D_{1})$ by weakly meet $s_{Z}$-continuity. Then $int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\cap\mathord{\downarrow}a\cap\mathord{\downarrow}D_{1}\neq\emptyset$, which implies that there is a $y_{1}\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\cap\mathord{\downarrow}a\cap\mathord{\downarrow}D_{1}$. By $y_{1}\leq a$ and $a\in D_{2}^{\delta}$, we have $y_{1}\in D_{2}^{\delta}$. Similarly, we get that $y_{1}\in cl_{\sigma^{Z}(P)}(\mathord{\downarrow}y_{1}\cap\mathord{\downarrow}D_{2})$ and $int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\cap\mathord{\downarrow}y_{1}\cap\mathord{\downarrow}D_{2}\neq\emptyset$. So there is a $y_{2}\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\cap\mathord{\downarrow}y_{1}\cap\mathord{\downarrow}D_{2}$. By induction, we find $y_{n}\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\cap\mathord{\downarrow}y_{n-1}\cap\mathord{\downarrow}D_{n}$, where $y_{0}=a$, clearly, $y_{n}\in\bigcap_{i=1}^{i=n}\mathord{\downarrow}D_{i}$. Since $y_{n}\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\subseteq\mathord{\uparrow}F$, $y_{n}\geq x_{i_{0}}$ for some $i_{0}\in\{1,2,...,n\}$, this implies $x_{i_{0}}\in\mathord{\downarrow}D_{i_{0}}$, which contradicts that $x_{i_{0}}\notin\mathord{\downarrow}D_{i_{0}}$. Hence, $int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\subseteq\cup\{\rotatebox{-90.0}{$\twoheadleftarrow$}_{Z}x:x\in F\}$.

By Lemma 3.8 and Lemma 3.10, obviously, $\Uparrow_{Z}F\subseteq\rotatebox{-90.0}{$\twoheadleftarrow$}_{Z}F$. And the reverse containment is easy to verify, so we omit the proof.

It suffices to prove that $x\in(\rotatebox{90.0}{$\twoheadleftarrow$}_{Z}x)^{\delta}$ for any $x\in P$. Suppose that there is a $y\in(\rotatebox{90.0}{$\twoheadleftarrow$}_{z}x)^{u}$ but $x\nleq y$. Then there are $U\in\sigma^{Z}(P)$, $V=P\setminus\mathord{\uparrow}F\in\omega(P)$ such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$, so $U\subseteq\mathord{\uparrow}F$. Since $\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\in\sigma^{Z}(P)$ by Lemma 3.2 and $x\in\mathord{\uparrow}(\mathord{\downarrow}x\cap U)\subseteq\mathord{\uparrow}F$, we have $x\in int_{\sigma^{Z}(P)}(\mathord{\uparrow}F)\subseteq\rotatebox{-90.0}{$\twoheadleftarrow$}_{Z}F$. Thus there is an $m\in F$ such that $m\in\rotatebox{90.0}{$\twoheadleftarrow$}_{Z}x$. It follows that $m\leq y$, then $y\in\mathord{\uparrow}F$. But this contradicts that $y\in V$.

$(1)\Rightarrow(2)$ Straightforward by Proposition 3.13 and Proposition 3.14.

$(2)\Rightarrow(3)$ For any $x\nleq y$, that is, $y\notin\mathord{\uparrow}x$, there is an $F\in\omega_{Z}(x)$ such that $y\notin\mathord{\uparrow}F$ by (2). So we get that there are $\Uparrow_{Z}F\in\sigma^{Z}(P)$, $P\setminus\mathord{\uparrow}F\in\omega(P)$ containing $x$ and $y$, respectively, and $\Uparrow_{Z}F\cap P\setminus\mathord{\uparrow}F=\emptyset$.

$(3)\Rightarrow(1)$ By Proposition 3.15 and $\rotatebox{90.0}{$\twoheadleftarrow$}_{Z}x\in I_{Z}(P)$, we know $P$ is $s_{Z}$-continuous.

Straightforward.

It is easy to verify that $(1)\Rightarrow(2)\Rightarrow(3)$.

$(3)\Rightarrow(4):$ It is clear that $D^{\delta}\subseteq D^{\delta}\mid_{A}$. Assume $m\in D^{u}\mid_{A}$. Then $D\subseteq\mathord{\downarrow}m$ and $m\in A$. Since $Z$ is subset hereditary, $D\in Z(\mathord{\downarrow}m)$. Thus we have $D^{\delta}\mid_{m}=D^{\delta}$ by $(3)$. Now we only need to prove that $D^{\delta}\mid_{A}\subseteq D^{\delta}\mid_{m}$. Assume $a\in D^{\delta}\mid_{A}$, $b\in D^{u}\mid_{m}$. Then $b\leq m$ and $b\in A$ as $m\in A$, which implies that $b\in D^{u}\mid_{A}$, so $a\leq b$. Hence, $D^{\delta}\mid_{A}\subseteq D^{\delta}\mid_{m}$.

$(4)\Rightarrow(1):$ We want to prove that $\Gamma^{Z}(A)=\{A\cap C:C\in\Gamma^{Z}(P)\}$ for any $A\in\Gamma^{Z}(P)$. For each $B\in\Gamma^{Z}(A)$, let $D\in Z(P)$ and $D\subseteq B$. Then $D\in Z(A)$ because $Z$ is subset hereditary. It follows that $D^{\delta}\mid_{A}\subseteq B$, which means $D^{\delta}\subseteq B$ since $D^{\delta}\mid_{A}=D^{\delta}$. Thus $B\in\Gamma^{Z}(P)$ and $\Gamma^{Z}(A)\subseteq\{A\cap C:C\in\Gamma^{Z}(P)\}$. Conversely, for any $C\in\Gamma^{Z}(P)$, let $D\in Z(A)$ with $D\subseteq A\cap C$. Then $D\in Z(P)$ and $D^{\delta}\subseteq A\cap C$ since $A\cap C\in\Gamma^{Z}(P)$. This implies that $D^{\delta}\mid_{A}\subseteq A\cap C$. So $A\cap C\in\Gamma^{Z}(A)$, and hence, $\Gamma^{Z}(A)=\{A\cap C:C\in\Gamma^{Z}(P)\}$ holds.