How's it possible for a bounded subset of the reals not to contain it's $\inf$ or $\sup$? 
Definition
By the real numbers we mean $\Bbb R:=\{\gamma:\gamma \text{ is a cut of $\Bbb Q$}\}$.

In Principles of Mathematical Analysis Rudin proves that if $A\subseteq \Bbb R$ is bounded, and we let $$\alpha :=\bigcup_{\gamma \in A}\gamma$$
Then $\alpha = \sup A$.
From this construction, I don't understand how could there be a set $A$ in which $\bigcup_{\gamma \in A}\gamma\not \in A$, that is : a set that doesn't contain it's supremum.
Could someone clarify my confusion?
E:
It seems I wasn't clear enough: I'm aware such sets exist, what I don't understand is how, with this construction, could $\alpha$ not be an element of $A$ (as I understand it, $\alpha$ is "the biggest cut": one of the $\gamma$s in $A$, am I wrong?).
E2:

Definition
By a cut $\beta$ we mean a proper subset of $\Bbb Q$ which satisfies:

*

*$p\in \beta$ and $q<p$ implies $q \in \beta$.

*If $p \in \beta$ then there exists a $r \in\beta$ such that $p<r$.


I don't have a proof or anything for my above "claim", I just thought Rudin's argument would imply that $\alpha\in A$: I imagine cuts (informally, of course) as intervals of rational numbers $(-\infty,r)$ where the $r$ is determined by some condition (for example $\{x:x^2<2\}$ determine $r=\sqrt 2$, etc).
So looking at it that way, it looks like that when I say "$\alpha=\bigcup_{\gamma\in A} \gamma$", the resulting set (the union) must be just a single one of those $\gamma$s, more clearly, it looks like (to me) that $\alpha=\gamma_0$ where $\gamma_0\in A$ and $\gamma \subseteq \gamma_0\forall \gamma \in A$. Maybe I'm just interpreting wrongly what a cut is.
 A: In general, when you take a union of the sets in a collection of sets,
you can easily end up with a new set that is not in the original collection.
I'll assume that when we take $\bigcup_{\gamma\in A} \gamma$, where each 
$\gamma$ is a "cut", we are actually referring to a union of the lower sets
in each cut; that is, each $\gamma$ is a set of rationals that does not
contain its own supremum.
Let $A$ be the countably infinite collection of Dedekind cuts named
$1/2, 2/3, 3/4, 4/5, \ldots$.
That is, $A = \left\{ 1 - \frac1n \mid n \in \mathbb N \right\}$.
Now let $q$ be any rational number less than $1$;
for a large enough $n$, $q < 1 - \frac1n$, so $q$ is a member of the
cut named $1 - \frac1n$, and therefore $q \in A$.
That is, $A$ is the set of all rational numbers less than $1$,
and therefore $A$ is just exactly the Dedekind cut for the real number $1$.
In short, $\sup A = 1$ and $\bigcup_{\gamma\in A} \gamma = 1$, 
but $1 \not\in A$.
A: I'll work under the assumption that a cut $\gamma$ is a subset of $\mathbb{Q}$ such that


*

*$\gamma\ne\emptyset$;

*for all $a,b\in\mathbb{Q}$, if $a<b$ and $b\in\gamma$, then $a\in\gamma$;

*for all $a\in\gamma$, there exists $b\in\gamma$ with $a<b$;

*there exists $c\in\mathbb{Q}$ such that $c\notin\gamma$.


The order relation is $\gamma<\delta$ if and only if $\gamma\subsetneq\delta$.
A subset $A$ of $\mathbb{R}$ is bounded if there exists $\varepsilon\in\mathbb{R}$ such that $\gamma<\varepsilon$, for every $\gamma\in A$.
In particular, if $A$ is bounded, then $\delta=\bigcup_{\gamma\in A}\gamma\ne\mathbb{Q}$, because by definition, $\gamma\subseteq\varepsilon$, for every $\gamma\in A$ and so $\delta$ is a subset of $\varepsilon$.
Proving properties 1–3 of $\delta$ is straighforward, so $\delta$ is indeed a cut and, of course $\gamma\le\delta$, for every $\gamma\in A$.
Denote by $\mathbf{0}$ the cut
$$
\mathbf{0}=\{a\in\mathbb{Q}:a<0\}
$$
and consider $A=\{\gamma\in\mathbb{R}:\gamma<\mathbf{0}\}$. In particular, $\mathbf{0}\notin A$. Let's look at
$$
\delta=\bigcup_{\gamma\in A}\gamma
$$
Since every $\gamma\in A$ consists of negative rational numbers, also $\delta$ consists of negative rational numbers; in particular $\delta\subseteq\mathbf{0}$. Let $c\in\mathbb{Q}$, $c<0$. Then $c/2<0$ and
$$
\gamma_0=\{a\in\mathbb{Q}:a<c/2\}
$$
is a cut; moreover $\gamma_0<\mathbf{0}$ and $c\in\gamma_0$, so $c\in\delta$. Thus every negative rational number belongs to $\delta$ and we have so proved that $\delta=\mathbf{0}$.
This provides the required counterexample.
A: Hint: (0,1), or any bounded open set.
