# Least upper bound not an element in the cut

In his proof "Any bounded nonempty subset $\mathbb{R}$ has a least upper bound in $\mathbb{R}$", Enderton showed that the least upper bound is just $\bigcup A$. Since by definition $x\subseteq \bigcup A\quad \& \quad x\in A$ . In the case $x=\bigcup A\quad \& \quad x\in A$, $x$ becomes the largest element in $A$, but this cannot be true since the Dedekind Cut is a subset of $\mathbb{Q}$ that must have no largest element. How do you explain that?

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$A$ is a set of sets of rational numbers. While Dedekind cuts have no upper bound, there might very well be an upper bound to a set of Dedekind cuts.
Note that $x,y\in A$ the $x\subseteq y$ or $y\subseteq x$. Then $\bigcup A=\{z\in\mathbb Q\mid \exists x\in A(z\in x)\}$. That for itself is a Dedekind cut.
In particular, $\bar A=A\cup\{\bigcup A\}$ is a set of Dedekind cuts, and it is simple to see that every $x\in A$ is a subset of $\bigcup A$, as well $\bigcup A$ is a subset of itself. Therefore ordered by inclusion, $\bar A$ has a maximal element.