I'm trying to understand the following construction of an initial algebra from a weakly initial algebra (see here, section 4.5)
Let $\mathcal{A} = (A,s)$ be an algebra.
Let $M_s$ be the intersection of all sets $B$ such that $(B,s)$ is an algebra.
Let $M(\mathcal{A})$ be the minimal subalgebra of $\mathcal{A}$, then:
Statement 1: $\mathcal{M}(\mathcal{A}) = (M_s,s)$
Statement 2: there exists at most one morphism from $M(\mathcal{A})$ to any other algebra...
Statement 3: given a weakly initial algebra $\mathcal{C}$, the desired initial algebra is its minimal subalgebra $\mathcal{M}(\mathcal{C})$
I'm not understanding the construction well. How should I define minimal subalgebra? Why should that be equal to the intersection subalgebra $M_s$ with the same structure mapping?
I hope someone can clarify me these notions. I don't see this construction in the literature.
Update:
Section 7 of this paper seems to provide a better understanding.
As far as I can infer from the passage in the text
"Given an algebra $\mathcal{A}=(A,s)$, let $M_s$ be the intersection of all sets $B$ such that $(B,s)$ is an algebra and let $\mathcal{M}(\mathcal{A})$, the minimal subalgebra of $\mathcal{A}$, be $(M_s,s)$."
the minimal algebra $\mathcal{M}(\mathcal{A})$ is defined via this construction.
Observe that, since $(A,s)$ itself is an algebra, the set $A$ belongs to the collection of those sets that are used for the intersection, so one might as well restrict the intersection to all subsets of $A$.
A mathematician would probably just have written
"Given an algebra $\mathcal{A}=(A,s)$, let $\mathcal{M}(\mathcal{A})=(M_s,s)$ be the intersection of all subalgebras of $\mathcal{A}$"
but computer scientist sometimes prefer it a bit more convoluted."