TITLE: A curiosity: how do we prove $\mathbb{R}$ is closed under addition and multiplication?
QUESTION [19 upvotes]: So I tried looking around for this question, but I didn't find much of anything - mostly unrelated-but-similarly-worded stuff. So either I suck at Googling or whatever but I'll get to the point.
So far in my coursework, it seems like we've mostly taken for granted that $(\mathbb{R},+,\cdot)$ is a field. I'm not doubting that much, that would seem silly. However, my question is: how would one prove this? In particular, how would one prove that $(\mathbb{R},+)$ and $(\mathbb{R}\setminus \{0\}, \cdot)$ are closed under their respective operations?
I understand the definition of closure, but to say "a real number plus/times a real number is a real number" seems oddly circular since, without demonstrating that, it essentially invokes the assumptions we're trying to prove. Obviously, there's something "more" to the definition of "real number" that would make proving this possible.
Though I'm not sure what property would be used for this.
One thought I dwelled on for a while was instead looking at what the real numbers are not. For example, they are numbers lacking those "imaginary" components you see in their higher-dimensional generalizations - the complex numbers ($i$), quaternions ($i,j,k$), and so on. But that didn't seem quite "right" to me? Like I'm not sure if it's really wrong, it just irked me in some way. Like it's simple enough to say "a real number is any complex number with a zero imaginary component," take two real numbers, show their imaginary parts sum/multiply to zero, and thus we have a real number.
Maybe it's just a personal issue? Like I said - I'm not saying it's inherently wrong (it might be, though, I don't know - if it is, I would like to know why). Maybe it's just the whole idea of "defining a number by what's it's not" that bugs me. Like I said, I'm not really sure, and I think I'm rambling/unclear enough as it is, so I'll get straight to the point.

In short, how does one properly demonstrate, if not in the above way,
  $$a,b \in \mathbb{R} \Rightarrow (a+b)\in \mathbb{R}$$
$$a,b\in \mathbb{R \setminus \{0\}} \Rightarrow (a\cdot b) \in \mathbb{R \setminus \{0\}}$$

(And again, I'm not at all doubting that these are true. I'm just curious as to how one would demonstrate these facts in the most appropriate manner since I don't believe it's come up in my coursework and I've been curious on how one would prove it for a few days now.)

REPLY [6 votes]: Interestingly, all answers so far give Cauchy sequences, so let me give the Dedekind cut construction.
The development assumes that the rational numbers are known and constructed, as are their operations and basic properties. 
Definition. A cut is a partition $(A_1,A_2)$ of $\mathbb{Q}$, satisfying the following properties:

$A_1\neq\varnothing$ and $A_2\neq\varnothing$.
$A_1\cup A_2=\mathbb{Q}$.
If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
If there exists a rational $c\in\mathbb{Q}$ such that for every $a\in A_1$ and every $b\in A_2$, $a\leq c$ and $c\leq b$, then $c\in A_1$. 

In particular, if $a\in A_1$ and $c\lt a$, then $c\in A_1$ (since we must have $c\in A_1\cup A_2$).  Point 4 is somewhat arbitrary: you could also require this element to be in $A_2$. Or you could, as Dedekind does, simply ignore it and eventually identify some cuts together. Also, point 2 is somewhat superfluous, as it is included in the notion of "partition".
Let us call $A_1$ the "left set" of the cut, and $A_2$ the "right set".
Now, every rational number $r$ determines a cut, defined by $(A^r_1,A^r_2)$, with $A^r_1=\{q\in\mathbb{Q}\mid q\leq r\}$, and $A^r_2 = \{q\in\mathbb{Q}\mid q\gt r\}$. However, not every cut is determined by a rational: for example, the cut $(A_1,A_2)$ given by
$$\begin{align*}
A_1 &= \{q\in\mathbb{Q}\mid q\lt 0\} \cup \{q\in\mathbb{Q}\mid q\geq 0\text{ and }q^2\leq 2\}\\
A_2 &= \{q\in\mathbb{Q}\mid g\geq 0\text{ and }q^2\gt 2\}
\end{align*}$$
can be shown not to be of the form $(A_1^r,A_2^r)$ for any rational number $r$.
We define the set of real numbers $\mathbb{R}$ as the set of all cuts. 
We define the order in $\mathbb{R}$ by letting $(A_1,A_2)\leq (B_1,B_2)$ if and only if $A_1\subseteq B_1$. 
We define addition of cuts as follows: given $(A_1,A_2)$ and $(B_1,B_2)$, we define the cut $(A_1,A_2)\oplus (B_1,B_2) = (C_1,C_2)$, where $C_1$ consists of all rational numbers $q$ for which there exists $a\in A_1$ and $b\in B_1$ with $q\leq a+b$; and letting $C_2=\mathbb{Q}\setminus C_1$. 
It is not hard to verify that $-(A_1,A_2) = (-A_2,-A_1)$, where $-X = \{-q\mid q\in X\}$; except that you need to be a bit careful given my definition: to make sure it satisfies point 4, if there is a rational $c$ in $A_1$ Such that $q\leq c$ for all $q\in A_1$, then you should move $-c$ out of $-A_1$ and put it in $-A_2$.
Now, multiplication is a bit more complicated. First we let $\mathbf{0}=(0_L,0_R)$, where $0_L=\{q\in\mathbb{Q}\mid q\leq 0\}$, and $0_R=\{q\in\mathbb{Q}\mid q>0\}$. 
Then, given two cuts $(A_1,A_2)$, $(B_1,B_2)$, with both greater than or equal to $(0_L,0_R)$, we define their product $(A_1,A_2)\otimes(B_1,B_2) = (C_1,C_2)$ by letting 
$$C_1 = \{q\in\mathbb{Q}\mid \exists a\in A_1, b\in B_1\text{ such that }a\geq 0,b\geq 0,\text{ and }q\leq ab\}$$
and then lettig $C_2 = \mathbb{Q}\setminus C_1$. 
If $(A_1,A_2)\geq (0_L,0_R)$ and $(B_1,B_2)\lt (0_L,0_R)$, we define the product $(A_1,A_2)\otimes(B_1,B_2)$ to be $-\bigl( (A_1,A_2)\otimes(-B_2,-B_1)\bigr)$. 
And if $(A_1,A_2)\lt (0_L,0_R)$, and $(B_1,B_2)\lt(0_L,0_R)$, we define
$$(A_1,A_2)\otimes(B_1,B_2) = (-A_2,-A_1)\otimes(-B_2,-B_1).$$
One can then show that the map $\mathbb{Q}\to\mathbb{R}$ given by $r\mapsto (A_1^r,A_2^r)$ is one-to-one, and satisfies
$$\begin{align*}
(A_1^{r+s},A_2^{r+s}) &= (A_1^r,A_2^r) \oplus (A_1^s,A_2^s)\\
(A_1^{rs},A_2^{rs}) &= (A_1^r,A_2^r)\otimes (A_1^s,A_2^s)\\
r\leq s&\iff (A_1^r,A_2^r)\leq (A_1^s,A_2^s)
\end{align*}$$
so that $\mathbb{R}$ contains a "copy" of $\mathbb{Q}$ inside of it. Finally, one can show completeness by showing that if $\{(A^{(i)}_1,A^{(i)}_2)\}_{i\in I}$ is a nonempty family of real numbers that is bounded above (so there exists a real number $(B_1,B_2)$ such that $A^{(i)}_1\subseteq B_1$ for all $i$), then the real number $(A_1,A_2)$ with $A_1 = \cup_{i\in I}A^{(i)}_1$, $A_2=\mathbb{Q}-A_1$ is the supremum of the set.