I know that it is possible to represent every finite lattice $L$ with a union-closed family  $\mathcal{F}$ containing the empty set: for every $x\in L$, let $S_x=\{y\in L\, :\, y\not\geq x\}$ and $\mathcal{F} = \{S_x :\, x\in L\}$ (see [this question](https://mathoverflow.net/q/469011/136218)).

However, the number of elements in the universe $|U(\mathcal{F})| =|L|$ is not necessarily minimal: we can remove some of them and still have a family representing the lattice.

Here are a few examples where the removed (not needed) elements are depicted in red:

The powerset:

[![enter image description here][1]][1]

or:
[![enter image description here][2]][2]

the above slightly modified:
[![enter image description here][3]][3]

[this example family](https://mathoverflow.net/a/228124/136218):

[![enter image description here][4]][4]

I have done the above with a program that tries to remove one element at a time. It seems that the result does not depend on the order of tries.

Is it possible to identify systematically the elements that are not removed and/or determine or estimate the minimum number of elements needed for a given lattice based on some structure of it?

Maybe the only needed elements are those with out-degree exactly $1$?

EDIT: the following further examples seem to support the above hypothesis:

[this union-closed family](https://mathoverflow.net/a/162740/136218):

[![enter image description here][5]][5]

[the Tamari lattice of order 4](https://en.wikipedia.org/wiki/Tamari_lattice):

[![enter image description here][6]][6]

[the Stanley lattice N=4](https://houseofgraphs.org/graphs/33589)

[![enter image description here][7]][7]

Following the hint in the comment, let $\mathcal{F} = \{S_x :\, x\in L\}$ and $S_x=\{y\in L\, :\, y\not\geq x,\, y \not= a \land b\}$, i.e. $y$ must be meet-irreducible.

Then $S_t \cup S_u = \{y\in L\, :\, y\not\geq t \lor y\not\geq u,\, y \not= a \land b\} = \{y\in L\, :\, y\not\geq t \lor u,\, y \not= a \land b\} = S_{t \lor u}$, because $y \ge t \lor u \iff y \ge t$ and $y \ge u$.

Now suppose $S_t = S_u$ for some $t \not= u$. Then any $y$ such that $y \ge t$ and $y \not\ge u$ must be meet-reducible, i.e. $y = a \land b$. But then $a \ge t$, $b \ge t$, $a \not\ge u$ or $b \not\ge u$. Then $a \ge t$ and $a \not\ge u$, or $b \ge t$ and $b \not\ge u$. Then $a$ or $b$ must be meet-reducible. But we cannot continue indefinitely, therefore $S_t = S_u \iff t = u$.

Let $t$ be meet-irreducible and $u$ the least element $x \gt y$, then $t \in S_u$ and $t \not\in S_t$. If $y \ge u$ then $y \ge t$. If we remove $t$ from the universe, since $t$ is meet-irreducible, then any $y \ge t$, $y \not= t$, must satisfy $y \ge u$. Therefore $S_t = S_u$ and we cannot remove any of the meet-irreducible elements.

Although $\mathcal{F}$ is a valid representation of the lattice, I think that it is not proved that the number of elements of its universe is minimal. Is it possible to prove or disprove it?

  [1]: https://i.sstatic.net/2f8LLPwM.png
  [2]: https://i.sstatic.net/7orB775e.png
  [3]: https://i.sstatic.net/kEyW33Jb.png
  [4]: https://i.sstatic.net/fItYNp6t.png
  [5]: https://i.sstatic.net/itZnYFlj.png
  [6]: https://i.sstatic.net/l7JDsU9F.png
  [7]: https://i.sstatic.net/ouP8BbA4.png