Domination problem with sets

An algorithmic solution:

Clearly we can assume that each element in $M$ appears in exactly $4$ subsets. Let $|M| =n$.

Stage 1. Take maximal subfamily $\mathcal{A} \subseteq \{S_1,....,S_k\} =:\mathcal{S} $ such that:

$\bullet$ every member of that family $\mathcal{A}$ has 3 elements;

$\bullet$ all sets in $\mathcal{A}$ are pairwise disjunct.

Let $|\mathcal{A}| =a$ and let $A= \cup _{X\in \mathcal{A}} X$. Then $|A|=3a$. Now, since each $a\in A$ appears exactly $4$ times it must appear exactly $3$ times in sets not in $\mathcal{A}$. So by double counting between $M$ and $\mathcal{S}\setminus \mathcal{A}$ we have (elements in $A$ have a degree $3$ and other $4$) $$ 3\cdot 3a +4(n-3a)\leq 3\cdot (k-a)\;\;\Longrightarrow \;\;4n \leq 3k\;\;\;...(1)$$

Let us now erase all the elements in $M$ which appears in $A$ and let this new set be $M_1$, so $M_1 = M\setminus A$ (so $|M_1| = n-3a$) and do the same thing in remaining sets in $\mathcal{S}\setminus \mathcal{A}$ and we get new family of sets $\mathcal{S}_1$.

Notice that each element in $M_1$ appears still $4$ times in sets from $\mathcal{S}_1$ and that each set in $\mathcal{S}_1$ has at most $2$ elements. (Why? If some of it, say $X$, has $3$ elements, that means that no element in $X$ was erased, so no element in $X$ is in $A$. But then we could put $X$ in $\mathcal{A}$ and we would get bigger family than $\mathcal{A}$ which is already maximal.) Also, let $k_1=|\mathcal{S}_1|$

Stage 2. Now take a maximal subfamily $\mathcal{B} \subseteq \mathcal{S}_1 $ such that:

$\bullet$ every member of that family $\mathcal{B}$ has 2 elements;

$\bullet$ all sets in $\mathcal{B}$ are pairwise disjunct.

Let $|\mathcal{B}| =b$ and let $B= \cup _{X \in \mathcal{B}} X$. Then $|B|=2b$. Now, since each $b\in B$ appears exactly $4$ times it must appear exactly $3$ times in sets not in $\mathcal{B}$. So by double counting between $M_1$ and $\mathcal{S}_1\setminus \mathcal{B}$ we have (elements in $B$ have a degree $3$ and other $4$) $$ 3\cdot 2b +4(n-3a-2b)\leq 2\cdot (k_1-b)\;\;\Longrightarrow \;\;2n \leq k+5a\;\;\;...(2)$$

Let us now erase all the elements in $M_1$ which appears in $B$ and let this new set be $M_2$, so $M_2 = M_1\setminus B$ (so $|M_2| =n-3a-2b$) and do the same thing in remaining sets in $\mathcal{S}_1\setminus \mathcal{B}$ and we get new family of sets $\mathcal{S}_2$.

Notice that each element in $M_2$ appears still 4 times in sets from $\mathcal{S}_2$ and that each set in $\mathcal{S}_2$ has at most 1 elements. (Why? If some of it, say $X$, has 2 elements, that means that no element in $X$ was erased, so no element in $X$ is in $B$. But then we could put $X$ in $\mathcal{B}$ and we would get bigger family than $\mathcal{B}$ which is already maximal.) Also, let $k_2=|\mathcal{S}_2|$.

Final stage. Now take a maximal subfamily $\mathcal{C} \subseteq \mathcal{S}_2 $ such that:

$\bullet$ every member of that family $\mathcal{C}$ has 1 element;

$\bullet$ all sets in $\mathcal{C}$ are pairwise disjunct.

Let $|\mathcal{C}| =c$ and let $C= \cup _{X \in \mathcal{C}} X$. Then $|C|=c$. Now, since each $c\in C$ appears exactly 4 times it must appear exactly 3 times in sets not in $\mathcal{C}$. So by double counting between $M_2$ and $\mathcal{S}_2\setminus \mathcal{C}$ we have (elements in $C$ have a degree $3$ and other $4$) $$ 3\cdot c +4(n-3a-2b-c)\leq 1\cdot (k_2-c)\;\;\Longrightarrow \;\;4n \leq k+11a+7b\;\;\;...(3)$$

Clearly $C=M_2$ so we are finish with the process, that is $c+2b+3a = |M| =n$. All we have to check if resurrected sets (that is, we refile all the sets with erased elements) satisfies $$a+b+c\leq {3\over 7}k\;\;\;\; {\bf ?}$$

Using (1) and $3a+2b+c=n$ we get: $$ 12a+8b+4c\leq 3k$$ Using (2) and $3a+2b+c=n$ we get: $$ a+4b+2c\leq k$$ Using (3) and $3a+2b+c=n$ we get: $$ 2a+2b+8c\leq 2k$$ If we add these three inequalites we get So $$ 14(a+b+c)<15a+14b+14c\leq 6k$$ and thus a conclusion.