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upper set
Let $P$ be a poset and $A$ a subset of $P$. The upper set of $A$ is defined to be the set
$\{b\in P\mid a\leq b\mbox{ for some }a\in A\},$ 
and is denoted by $\uparrow\!\!A$. In other words, $\uparrow\!\!A$ is the set of all upper bounds of elements of $A$.
$\uparrow$ can be viewed as a unary operator on the power set $2^{P}$ sending $A\in 2^{P}$ to $\uparrow\!\!A\in 2^{P}$. $\uparrow$ has the following properties
1. $\uparrow\!\!\varnothing=\varnothing$,
2. $A\subseteq\uparrow\!\!A$,
3. $\uparrow\uparrow\!\!A=\uparrow\!\!A$, and
4. if $A\subseteq B$, $\uparrow\!\!A\subseteq\uparrow\!\!B$.
So $\uparrow$ is a closure operator.
An upper set in $P$ is a subset $A$ such that its upper set is itself: $\uparrow\!\!A=A$. In other words, $A$ is closed with respect to $\leq$ in the sense that if $a\in A$ and $a\leq b$, then $b\in A$. An upper set is also said to be upper closed. For this reason, for any subset $A$ of $P$, the $\uparrow\!\!A$ is also called the upper closure of $A$.
Dually, the lower set (or lower closure) of $A$ is the set of all lower bounds of elements of $A$. The lower set of $A$ is denoted by $\downarrow\!\!A$. If the lower set of $A$ is $A$ itself, then $A$ is a called a lower set, or a lower closed set.
Remarks.

$\uparrow\!\!A$ is not the same as the set of upper bounds of $A$, commonly denoted by $A^{u}$, which is defined as the set $\{b\in P\mid a\leq b\mbox{ for \emph{all} }a\in A\}$. Similarly, $\downarrow\!\!A\neq A^{{\ell}}$ in general, where $A^{{\ell}}$ is the set of lower bounds of $A$.

When $A=\{x\}$, we write $\uparrow\!\!x$ for $\uparrow\!\!A$ and $\downarrow\!\!x$ for $\downarrow\!\!A$. $\uparrow\!\!x=\{x\}^{u}$ and $\downarrow\!\!x=\{x\}^{d}$.

If $P$ is a lattice and $x\in P$, then $\uparrow\!\!x$ is the principal filter generated by $x$, and $\downarrow\!\!x$ is the principal ideal generated by $x$.

If $A$ is a lower set of $P$, then its set complement $A^{{\complement}}$ is an upper set: if $a\in A^{{\complement}}$ and $a\leq b$, then $b\in A^{{\complement}}$ by a contrapositive argument.

Let $P$ be a poset. The set of all lower sets of $P$ is denoted by $\mathcal{O}(P)$. It is easy to see that $\mathcal{O}(P)$ is a poset (ordered by inclusion), and $\mathcal{O}(P)^{{\partial}}=\mathcal{O}(P^{{\partial}})$, where ${}^{{\partial}}$ is the dualization operation (meaning that $P^{{\partial}}$ is the dual poset of $P$).
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