Let $\mathcal{A}$ be a finite family of sets. For $\mathcal{A}' \subseteq \mathcal{A}$, define $\cup \mathcal{A}'=\bigcup_{A \in \mathcal{A}'}A$. Let $U(\mathcal{A})=\{\cup \mathcal{A}' : \mathcal{A}' \subseteq \mathcal{A}\}$, considered as a poset ordered by inclusion.

I'm trying to show that $U(\mathcal{A})$ is a lattice.

$U(\mathcal{A})$ is a lattice if every pair $x,y \in L$ has a unique largest common lower bound, called their $\textbf{meet}$, written $x \wedge y$, and also has a unique smallest common upper bound, called their $\textbf{join}$ and written $x \vee y$. That is to say, $\forall z \in U(\mathcal{A})$

$$z \leq x \text{ and } z \leq y \rightarrow z \leq x \wedge y$$

$$z \geq x \text{ and } z \geq y \rightarrow z \geq x \vee y$$

My thoughts:

At first I just wanted to say that the join of two elements would be their union, but then I realized that their union is not necessarily a well defined element.

Eventually I realized that this question wasn't asking to give an explicit construction of the meet and join, but rather just show that they exist. Furthermore, since $\mathcal{A}$ is a finite family of sets, the poset $U(\mathcal{A})$ is bounded, and therefore meets and joins will indeed exist, and I just need to show that they are unique.

If anyone could help me out with this that'd be great, I've been stuck on this for longer than I'd like to admit haha.


1 Answer 1


If $x_i\in U(\mathcal{A})$ for $i\in\{0,1\}$, then

$$x_0\lor x_1=\bigcup\{A\in\mathcal{A}:A\subseteq x_0\cup x_1\}\in U(\mathcal{A})\,,\tag{1}$$


$$x_0\land x_1=\bigcup\{A\in\mathcal{A}:A\subseteq x_0\cap x_1\}\in U(\mathcal{A})\,.$$

$(1)$ can actually be simplified. There are $\mathcal{A}_i\subseteq\mathcal{A}$ for $i\in\{0,1\}$ such that each $x_i=\bigcup\mathcal{A}_i$, and $(1)$ reduces to

$$x_0\lor x_1=x_0\cup x_1=\bigcup(\mathcal{A}_0\cup\mathcal{A}_1)\in U(\mathcal{A})\,.$$

  • $\begingroup$ Hmmm.. Do you have any reason why this exercise is giving with the following: hint: Don't try to specify the meet operation explicitly. $\endgroup$
    – user637978
    Jan 24, 2021 at 16:00
  • 1
    $\begingroup$ @ADragon: I can’t be certain, but I suspect that it’s to guide you away from trying $x_0\cap x_1$. That is, the join really is just a familiar operation on $x_0$ and $x_1$, but the meet can only be defined in a much more complicated way. I only included $(1)$ for the join in order to show that the two were really being defined similarly, so as to make the definition of the meet seem more natural. $\endgroup$ Jan 24, 2021 at 18:01
  • $\begingroup$ Thanks you're the bomb!! Is there a way we could have concluded the existence of the meet once we knew what the join was?? I'm trying to understand what other way of solving the problem the hint might have been guiding me towards... $\endgroup$
    – user637978
    Jan 24, 2021 at 18:07
  • 1
    $\begingroup$ @ADragon: You’re welcome! I don’t know; there might well be, but my knowledge of lattice theory is pretty elementary. $\endgroup$ Jan 24, 2021 at 18:14

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