Real numbers as element of a universe defn. A universe is a set $U$ such that:


*

*$x\in u\in U\Rightarrow x\in U$

*$u\in U$ and $v\in U$ imply $\left\{u,v\right\}, \langle u,v\rangle, u\times v\in U$

*$x\in U\Rightarrow \mathcal{P}(x)$ and $\cup x\in U$

*$\omega\in U$

*if $f:x\longrightarrow y$ is a surjective mapping with $x\in U$ and $y\subseteq U$ then $y\in U$
Questions:
1) is the set $\mathbb{R}$ of real numbers an element of $U$?
2) if yes, how can be proved?
 A: Well, it depends on what are the real numbers for you. Set theory and logic are very implementation agnostic when it comes to "everyday mathematics". Nobody cares what is the set $1_\Bbb R$, just that we can find a construct in which this set exists and behave as expected.
So for the sake of argument, let's take the usual construction:


*

*The natural numbers are the finite von Neumann ordinals.

*The integers are equivalence classes of natural numbers.

*The rational numbers are equivalence classes of integers.

*The real numbers are non-empty proper initial segments of the rational numbers.


So by assumption $\omega\in U$, so the natural numbers are there. Therefore the power set of $\omega$ is there, so the set of equivalence classes defining $\Bbb Z$ is there, similarly we get $\Bbb Q$ there, and finally we get $\Bbb R$.
So yes. A universe also contains the set of all the real numbers. But in reality we can say a lot more.

Theorem. If $U$ is a universe, then there exists a strongly inaccessible cardinal $\kappa$ such that $U=V_\kappa$. Moreover $U$ is a model of $\sf ZFC$. (Where $V_\kappa$ is the $\kappa$th stage of the von Neumann hierarchy.)

This is not a very difficult proof, and essentially was proved by Zermelo in the 1930's. But it shows that a universe is really big. 
A: $\mathbb{Z}$ is usually constructed is such a way that its elements
form a partition of $\omega\times\omega$.
So $\mathbb{Z}\subset\wp\left(\omega\times\omega\right)$ hence $\mathbb{Z}\in\wp\left(\wp\left(\omega\times\omega\right)\right)\in U$
and consequently $\mathbb{Z}\in U$.
$\mathbb{Q}$ is usually constructed is such a way that its elements
are disjoint subsets $\mathbb{Z}\times\left(\omega-\left\{ 0\right\} \right)$.
As above we find $\mathbb{Q}\in U$.
$\mathbb{R}$ can be constructed is such a way that its elements are
subsets of $\mathbb{Q}$.
So $\mathbb{R}\subset\wp\left(\mathbb{Q}\right)$ hence $\mathbb{R}\in\wp\left(\wp\left(\mathbb{Q}\right)\right)\in U$
and consequently $\mathbb{R}\in U$.
A: Let me outline a proof. As previously stated in a comment, the only nontrivial part is to actually define what $\Bbb R$ is. In set theory, it is common to just let $\Bbb R := \mathcal P(\omega)$ and in this case, we immediately have that $\Bbb R \in U$. However, this definition is not very common amongst non-set theorists, and I shall therefore outline a different approach, using Dedekind cuts (as drhab suggested).
First of all, let us define $\Bbb Z := \{ (a,b) \in \omega \times \omega \mid a = 0 \vee b = 0 \}$. We may think of $(a,b)$ as the integer $a-b$. Usually, the part "$a = 0 \vee b = 0$" is omitted and one simply defines an equivalence relation $\sim$ on $\omega \times \omega$ by letting $(a,b) \sim (c,d)$ iff $a+d = c+b$ (which in our indented interpretation just means that $a-b = c-d$. However, we can't write this down until we actually defined $-$. Note that in contrast $+$ is defined, as the usual ordinal addition.)
Clearly $\Bbb Z \subseteq \omega \times \omega$ and hence $\Bbb Z \in \mathcal P(\omega \times \omega)$. Since, $\omega, \omega \times \omega$ and $\mathcal P(\omega \times \omega)$ are elements of $U$, we now get that $\Bbb Z \in U$ as well, by condition 1.
Next, we define $\Bbb Q = \{ (p,q) \in \Bbb Z \times \omega \mid q \neq 0 \wedge (p,q) = 1 \}$. Here $(p,q) = 1$ means that $p$ and $q$ are coprime. I leave it to you to formalize this. (There are two cases, depending on whether $p = (a,0)$ or $p = (0,b)$ and both are trivial.) Again, one usually defines $\Bbb Q$ simply as $\Bbb Z \times (\omega \setminus \{0\})$ and takes the obvious quotient of this, as in the case of $\Bbb Z$. Repeating the argument above, we now get that $\Bbb Q \in U$.
This allows us to define $\Bbb R$: We say that $C \subseteq \Bbb Q$ is a cut  iff $C \not \in \{ \emptyset, \Bbb Q \}$, $C$ is downward closed, i.e. for every $p \in C$ and every $q \in \Bbb Q$ with $q < p$, we have that $q \in C$, and $C$ has no $<$-maximal element. (Note that we didn't define the order $(\Bbb Q; <)$ and we actually don't need to do this - formally. We may just work with our intuitive notion of $<$ on $\Bbb Q$. It is, however, not difficult to define $<$ as a set and to show that ${<} \in U$.) We now let $\Bbb R := \{ C \subseteq \Bbb Q \mid C \text{ is a cut} \}$.Our intended interpretation is that $C$ represents that unique real $x$ such that $\{ y \in \Bbb Q \mid y < x \} = C$. Since $\mathbb R \in \mathcal P(\mathcal P(\Bbb Q))$ and $\Bbb Q \in U$, we may apply condition 1 and 3 to conlude that $\Bbb R \in U$.
So far, we only defined a set $\Bbb R$ and it is not entirely obvious that this set has anything in common with our intuitive definition of $\Bbb R$. It is however not difficult to define sets $+, {\,\cdot\,}, {<} \in U$ such that $(\Bbb R, +, {\,\cdot\,}, {<}) \in U$ is provably isomorphic to our intuitive notion of $\Bbb R$ in the sense that $(\Bbb R, +, {\,\cdot\,}, {<})$ is a complete, totally ordered field.
