From "Universal Algebra: Fundamentals amd Selected Topics" by Clifford Bergman, "Let L be a lattice. Prove that for any subset X of L, the ideal generated by X is $$Ig^L(X)=\{a \in L\mid a\leq \bigvee Y, some Y \subseteq_{\mathbb{N}} X\}$$ Conclude that Idl(L) is an algebraic lattice."
Here we define Idl(L) as the set of ideals of L. A nonempty subset I of L is an ideal if I is a downset that is closed under finite joins. I have already proven that (Idl(L), $\subseteq$) is a complete lattice with I$\vee$J=I+J=$\{x+y\mid x\in I, y\in J\}$.
How do I prove that Idl(L) is an algebraic lattice? A lattice is algebraic if every element is the join of compact elements. How do I prove every ideal is the join (sum) of compact ideals?
Let us see that a principal ideal is compact.
Given $a \in L$, suppose $(a]$, the principal ideal generated by $a$ can be written as a join of ideals $(K_j)_{j\in J}$ of $L$ (where $J$ is an idea set).
It then follows that $a \in \bigvee_{j\in J}K_j$, whence there exist $a_1, \ldots, a_m \in \bigcup_{j \in J}K_j$ such that $a \leq a_1 \vee \cdots \vee a_m$.
Now, given that $a_1, \ldots, a_m \in \bigcup_{j \in J}K_j$, there exist $j_1, \ldots, j_m \in J$ such that $a_1 \in K_{j_1}, \ldots, a_m \in K_{j_m}$.
Hence, $a \in K_{j_1} \vee \cdots \vee K_{j_m}$, and $(a] \subseteq K_{j_1} \vee \cdots \vee K_{j_m}$; the reverse inclusion is obvious, by the definition, and so $(a] = K_{j_1} \vee \cdots \vee K_{j_m}$.
We conclude that $(a]$ is compact.
(We could proceed to show that non-principal ideals are not compact, but that is beyond the point here.)
Now we show that every ideal is the join of compact ideals, by showing that it is the join of principal ideals.
Take any ideal $I$ of $L$. If $b \in I$, then $b \in (b] \subseteq \bigcup_{a \in I}(a]$. Thus $I \subseteq \bigcup_{a \in I}(a]$.
It follows that $$I \subseteq \bigcup_{a\in I}(a] \subseteq \bigvee_{a \in I}(a] \subseteq I,$$ where the second inclusion in the above displayed formula follows from the fact that in a join, each joinand is less than the join (here, in the lattice of ideals, $(b] \leq \bigvee_{a \in I}(a]$, for each $b \in I$), and the third from the fact that if $a \in I$, then $(a] \subseteq I$, and if for all $a \in I$ we have $(a] \subseteq I$, then $\bigvee_{a\in I}(a] \subseteq I$ (it's just the definition of supremum).
Given that sequence of inclusions (starting and ending in $I$), we conclude that they're all equalities, and therefore $I = \bigvee_{a\in I}(a]$, a join of compact ideals.
As $I$ was any ideal, we conclude that the lattice of ideals of $L$ is algebraic.