In "Notes on Lattice Theory" J. B., there is a theorem such that
Let $\Gamma$ be a closure operator on a set $X$. The following are equivalent.
(1) $\Gamma$ is an algebraic closure operator.
(2) The union of any up-directed set of $\Gamma$-closed sets is $\Gamma$-closed.
(3) The union of any chain of $\Gamma$-closed sets is $\Gamma$-closed.
and A subset $D$ of an ordered set $P$ is said to be up-directed if for every $x,y \in D$ there exists $z \in D$ with $x \leq z$ and $y \leq z$.
The definition of algebraic closure operator
A closure operator $\Gamma$ on a set $X$ is said to be algebraic if for every $B \subseteq X$, $$\Gamma(B) = \bigcup \{\Gamma(F): F\ \text{is a finite subset of}\ B\}.$$
How to proof this theorem? like (1) $\Rightarrow$ (2), i cant figure out whats the up-directed set of closed set here.
For 1 to 2, let $\mathscr{C}$ be a an up-directed family of $\Gamma$-closed sets. (This means that $\Gamma(C)=C$ for all $C \in \mathscr{C}$ and $C_1,C_2 \in \mathscr{C}$ implies that there is a $C_3 \in \mathscr{C}$ that contains them both (the up-directed part). Let $\hat{C}= \bigcup \mathscr{C}$ and we have to show $\hat{C}$ is closed.
Now take any finite $F \subseteq \hat{C}$. So for any $f \in F$ we have some $C_f \in \mathscr{C}$ so that $f \in C_f$. Now we apply the up-directedness (which we can extend to finitely many sets by induction) and find some $C_0 \in \mathscr{C}$ such that $C_f \subseteq C_0$ for all $f \in F$ and so $F \subseteq C_0$. It follows that $\Gamma(F) \subseteq \Gamma(C_0) = C_0 \subseteq \hat{C}$ (by $\Gamma$-closedness of $C_0$ and monotonicity of $\Gamma$ under inclusion).
So this holds for all finite subsets of $\hat{C}$ and now using the fact that $\Gamma$ is an algebraic closure:
As $$\Gamma(\hat{C}) = \bigcup \{ \Gamma(F)\mid F \subseteq \hat{C} \text{ finite }\}\} \subseteq \hat{C} \subseteq \Gamma(\hat{C})$$
so $\hat{C}$ is $\Gamma$-closed.
(2) implies (3) is trivial as a chain of subsets is an up-directed family.