Trying to prove that the poset $U(\mathcal{A})$ is a lattice

Question

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, $\forall z \in U(\mathcal{A})$

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

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

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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.

Answer 1

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

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

and

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

$(1)$ 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

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