A_{\alpha}$, it follows that $\bigcup A_{\alpha}$ is an upper bound of $\mathcal{A}$; this is also be the smallest of all such bounds because if $U$ is any other upper bound then every $A_{\alpha}$ must precede $U$ by Eq. (\[Eqn: upper bound\]) and therefore so must $\bigcup A_{\alpha}$ (because the union of a class of subsets of a set is the smallest that contain each member of the class: $A_{\alpha}\subseteq U\Rightarrow\bigcup A_{\alpha}\subseteq U$ for subsets $(A_{\alpha})$ and $U$ of $X$). Analogously, since $\bigcap A_{\alpha}$ is $\subseteq$-less than each $A_{\alpha}$ it is a lower bound of $\mathcal{A}$; that it is the greatest of all the lower bounds $L$ in $\mathcal{X}$ follows because the intersection of a class of subsets is the largest that is contained in each of the subsets: $L\subseteq A_{\alpha}\Rightarrow L\subseteq\bigcap A_{\alpha}$ for subsets $L$ and $(A_{\alpha})$ of $X$. Hence the supremum and infimum of $\mathcal{A}$ in $(\mathcal{X},\subseteq)$ given by $$A_{\leftarrow}=\sup_{(\mathcal{X},\subseteq)}\mathcal{A}=\bigcup_{A\in\mathcal{A}}A\qquad\textrm{and}\qquad_{\rightarrow}A=\inf_{(\mathcal{X},\subseteq)}\mathcal{A}=\bigcap_{A\in\mathcal{A}}A\label{Eqn: supinf3}$$ are both elements of $(\mathcal{X},\subseteq)$. Intuitively, an upper (respectively, lower) bound of $\mathcal{A}$ in $\mathcal{X}$ is any subset of $\mathcal{X}$ that contains (respectively, is contained in) every member of $\mathcal{A}$.$\qquad\blacksquare$

The statement of Zorn’s lemma and its proof can now be completed in three stages as follows. For Theorem 4.1 below that constitutes the most significant technical first stage, let $g$ be a function on $(X,\preceq)$ that assigns to every $x\in X$ an *immediate successor* $y\in X$ such that $${\textstyle \mathscr{M}(x)=\{\textrm{ }y\succ x\!:\not\exists\textrm{ }x_{*}\in X\textrm{ satisfying }x\prec x_{*}\prec y\}}$$ are all the successors of $x$ in $X$ with no element of $X$ lying strictly between $x$ and $y$. Select a representative of $\mathscr{M}(x)$ by a choice fu