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 &quot;the biggest cut&quot;: 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 $q&lt;p$ implies $q \in \beta$.
If $p \in \beta$ then there exists a $r \in\beta$ such that $p&lt;r$.


I don't have a proof or anything for my above &quot;claim&quot;, 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 $\{x:x^2&lt;2\}$ determine $r=\sqrt 2$, etc).
So looking at it that way, it looks like that when I say &quot;$\alpha=\bigcup_{\gamma\in A} \gamma$&quot;, 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.
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 $n$, $q &lt; 1 - \frac1n$, 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 $1 \not\in A$.
Hint: (0,1), or any bounded open set.