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]


  [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