Lower bound for the size of a family of sets

Question

Consider a family $\mathcal{G} = \{ A_1,B_1,\ldots,B_m \}$ of $m+1$ non-empty finite distinct sets with the following property:

$$A_1 \cap B_k = \emptyset, 1 \le k \le m$$

Let $\mathcal{F} = \{A_1 \cup B_k, 1 \le k \le m\}$. Clearly $\mathcal{G}$ and $\mathcal{F}$ are disjoint. Also, $A_1 \cup B_j = A_1 \cup B_k$ implies $B_j = B_k$, therefore we know that $|\mathcal{F}| = m$.

Now let $\mathcal{G'} = \{ A_1',A_2',B_1',\ldots,B_m' \}$ be a family of $m + 2$ non-empty finite distinct sets with the following properties:

1. $A_1' \cap A_2' \cap B_k' = \emptyset, 1 \le k \le m$
2. $A_1' \not\subseteq A_2'$, $A_2' \not\subseteq A_1'$, $A_j' \not\subseteq B_k'$, $B_k' \not\subseteq A_j'$, $1 \le j \le 2$, $1 \le k \le m$

Let $\mathcal{F'} = \{A_1' \cup B_k', A_2' \cup B_k', A_1' \cup A_2' \cup B_k', 1 \le k \le m\}$. Clearly $\mathcal{G'}$ and $\mathcal{F'}$ are disjoint.

Any idea for extending the reasoning for $\mathcal{G}, \mathcal{F}$ to $\mathcal{G'}, \mathcal{F'}$ in order to obtain a lower bound for $|\mathcal{F'}|$ as a function of $m$?

For a motivation, I am trying to find a proof for case $m = 3$ in this question ($m$ there has not the same meaning as $m$ here).

EDIT:

There is a given answer that suggests that the lower bound is something near to $2 \sqrt{m}$. The first paragraph is clear to me, it gives an example where $|\mathcal{F}'| = 2\sqrt{m}+1$. From what I understand, the second paragraph says that the example given in the first paragraph is quite near to the actual lower bound, therefore I presume that the lower bound is $2 \sqrt{m}$ or something like that.

I understand that if e.g. $B_1' \cup A_i' = B_2' \cup A_1'$ for $i = 1$ or $i = 2$, then $B_1' \setminus B_2' \subseteq A_1'$. If we have also $B_1' \cup A_i' = B_2' \cup A_2'$ for $i = 1$ or $i = 2$, then $B_1' \setminus B_2' \subseteq A_2'$. Then $B_1' \setminus B_2' \subseteq A_1' \cap A_2'$, and $B_1' \setminus B_2' \subseteq A_1' \cap A_2' \cap B_1'$, but by hypothesis $A_1' \cap A_2' \cap B_1' = \emptyset$, therefore $B_1' \subseteq B_2'$. Then if we have also that $B_2' \cup A_i' = B_1' \cup A_1'$ for $i = 1$ or $i = 2$, and $B_2' \cup A_i' = B_1' \cup A_2'$ for $i = 1$ or $i = 2$, then $B_2' \subseteq B_1'$, then $B_1' = B_2'$, but $B_1', B_2'$ are required to be distinct.

It is enough to have $B_1' \cup A_1' = B_2' \cup A_1'$ and $B_1' \cup A_2' = B_2' \cup A_2'$ to imply $B_1' = B_2'$, or alternatively $B_1' \cup A_1' = B_2' \cup A_2'$ and $B_1' \cup A_2' = B_2' \cup A_1'$.

Getting the lower bound from there doesn't seem to be completely obvious.

Someone can help to elaborate the given answer?

Answer 1

Assume that $m=k^2$. Set $A_1’=\{p_1,\dots,p_k\}$, $A_2’=\{q_1,\dots, q_k\}$, and put $B_{i,j}’=A_1’\cup A_2’\setminus \{p_i,q_j\}$ (so there are $k^2$ sets of the form $B_{i,j}’$). Then $|\mathcal F’|$ is as small as $2k+1=2\sqrt m+1$.

This is close to the sharp lower bound; that can be shown after noticing that, if $B_i’\cup A_s’=B_j’\cup A_s’$ for $s=1,2$, then $i=j$.

Answer 2

Let $n=|\mathcal F'|$. Below we easily show that $n\ge\sqrt{m}$ and I shall look for more refined arguments to improve this bound.

For each natural $N$ put $[N]=\{1,2,\dots,N\}$. For each $i\in [2]$ put $\mathcal{F'_i}=\{A_i' \cup B_k':k\in [m]\}$ and $n_i=|\mathcal{F'_i}|$. Since $\mathcal{F'_i}\subset \mathcal{F'}$, $n_i\le n$. It is easy to see that for any $i\in [2]$ and $k,k'\in [m]$, $A'_i\cup B'_k=A'_i\cup B'_{k'}$ iff the symmetric difference $D=B'_k\Delta B'_{k'}$ is contained in $A'_i$. Therefore if both $A'_1\cup B'_k=A'_1\cup B'_{k'}$ and $A'_2\cup B'_k=A'_2\cup B'_{k'}$ then $D\subset A_1'\cap A_2'\cap (B_k'\cup B'_{k'})=\varnothing$ by Condition 1. Therefore the map $[m]\to \mathcal{F'_1}\times \mathcal{F'_2}$, $k\to (A_1'\cup B'_k,A_2'\cup B'_k)$ is injective. So $m=|[m]|\le n_1\times n_2\le n^2$.