A curiosity: how do we prove $\mathbb{R}$ is closed under addition and multiplication?

Let us assume that $\mathbb{Q}$ is defined in such a way that $(\mathbb{Q},+,\cdot)$ is a commutative ring without doubts.
Let us assume that $(a,b\in\mathbb{Q},a\neq b\Rightarrow a>b\vee a<b)$ is clear too. Let us define $\mathbb{R}$ as the metric completion of $\mathbb{Q}$.

Definition 1 - Cauchy convergence. We say that a sequence $\{q_n\}_{n\geq 1}$ of rational numbers is a Cauchy sequence if for any $\varepsilon>0$ there is a natural number $N=N(\varepsilon)$ ensuring $|q_n-q_m|\leq\varepsilon$ as soon as $n,m\geq N$. In layman's terms: a Cauchy sequence is a sequence whose terms "stick together" arbitrarily close from some point on.

Definition 2 - (metric) convergence. We say that a sequence $\{q_n\}_{n\geq 1}$ is (metrically) convergent to $q$ if for any $\varepsilon>0$ there is a natural number $N=N(\varepsilon)$ ensuring $|q_n-q|\leq\varepsilon$ as soon as $n\geq N$. In layman's terms: a convergent sequence is a sequence whose terms "stick arbitrarily close to something" from some point on.

Since not every Cauchy sequence in $\mathbb{Q}$ is metrically convergent to a rational number (you may take $q_n=\frac{F_{n+1}}{F_n}$ as an example), it makes sense to enrich $\mathbb{Q}$ in order to make every Cauchy sequence (in the new space) convergent (to something belonging to the new space). Let us consider the set $S$ of Cauchy sequences in $\mathbb{Q}$, and let us equip this space with an equivalence relation $\sim$ considering $\{q_n\}_{n\geq 1}$ and $\{p_n\}_{n\geq 1}$ as the same object when $\{p_n-q_n\}_{n\geq 1}$ is (metrically) convergent to zero.

$$\mathbb{R}\stackrel{\text{Def}}{=} S/\sim.$$ Since $p_n\to p$ (shorthand notation for "$\{p_n\}_{n\geq 1}$ is metrically convergent to $p$") and $q_n\to q$ imply $p_n+q_n\to p+q$ and $p_n\cdot q_n\to pq$, the ring structure of $\mathbb{Q}$ is inherited by $\mathbb{R}$ through metric completion.

In other terms: continuous maps ($+,\cdot$) defined over dense subsets ($\mathbb{Q}\subset\mathbb{R}$) have unique continuous extensions.

I believe that students' issues with complex numbers, in the majority of cases, can be easily explained. At least in my place we are accustomed to introducing the imaginary unit $i$ as a/"the" square root of $-1$, without properly justifying why we are allowed to introduce such monstrosity among numbers. In fact, the construction of $\mathbb{C}$ from $\mathbb{R}$ shares many elements with the construction of $\mathbb{R}$ from $\mathbb{Q}$:

1. We take the original space, $\mathbb{R}$
2. We take a huge superset, $\mathbb{R}[x]$, the ring of polynomials with real coefficients
3. We put an equivalence relation over such superset and define the new space as a quotient: $\mathbb{C}=\mathbb{R}[x]/(x^2+1)$
4. We check the new space is a field and a vector space of dimension $2$ over $\mathbb{R}$
5. We prove the new space is algebraically closed.

Or, at least, we should.

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:

1. $A_1\neq\varnothing$ and $A_2\neq\varnothing$.
2. $A_1\cup A_2=\mathbb{Q}$.
3. If $a\in A_1$ and $b\in A_2$, then $a\lt b$.
4. 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.

In the interest of disclosure:

This follows up on some comments made by Noah Schweber and Rahul to the original question, which were essentially nudges in this direction. This could also be taken as the "exercise left to the reader" bit of Xander Henderson's answer.

Essentially, since $a,b$ are real numbers, we let $a,b$ be represented by Cauchy sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$. If we can show $(a_n + b_n)$ and $(a_n b_n)$ are Cauchy, then the numbers they represent - $a+b$ and $a \cdot b$ respectively - are thus real numbers.

Ergo, the proof of closure under each operation is analogous to showing that, for two Cauchy sequences, their sum and products also are Cauchy.

I certainly don't intend to "accept" my answer here - in my opinion, it doesn't feel right in this case since it's basically just a "fleshing out" of what others told me, and the other answers here are also really good. But at the same time I'm posting this because I think it'll also be useful in case anyone else is looking in the future and may be confused on how to follow up on this idea. (Plus, I was already halfway through writing it before other people started posting their answers and didn't want my work to go to waste.)

Closure of Addition:

Okay, so let $(a_n), (b_n)$ be Cauchy sequences in $\mathbb{R}$, representing the real numbers $a$ and $b$ respectively. With this mind, we seek to prove that the sequence $(a_n + b_n)$, which represents, $a+b$, is Cauchy, and thus $(a+b)\in\mathbb{R}$.

Since $(a_n),(b_n)$ are Cauchy, for some $N_1, N_2 \in \mathbb{N}$

• $|a_m - a_n| < \varepsilon \;\;\; \forall \varepsilon > 0, \forall m,n > N_1$
• $|b_p - b_q| < \varepsilon \;\;\; \forall \varepsilon > 0, \forall p,q > N_2$

Since the inequalities hold for all $\varepsilon > 0$, we can in particular let them hold for $\varepsilon/2$. We will further let $N =\max\{N_1, N_2\}$.

Then we consider the condition for $(a_n + b_n)$ to be Cauchy and note:

$$|(a_m + b_m) - (a_n + b_n) | = |(a_m - a_n) + (b_m - b_n) |$$

Invoking the triangle inequality, we note, for all $m,n > N$ and $\varepsilon > 0$,

$$|(a_m - a_n) + (b_m - b_n) | \leq | a_m - a_n | + | b_m - b_n | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

Thus, since, for all $m,n > N =\max\{N_1, N_2\}$, we have $$|(a_m + b_m) - (a_n + b_n) |< \varepsilon$$ the sequence $(a_n + b_n)$ is Cauchy.

Since $(a_n + b_n)$ is Cauchy and represents $a+b$, we have $a+b \in \mathbb{R}$, which followed from our assumption $a,b \in \mathbb{R}$ (which gave us the assumptions $(a_n),(b_n)$ were Cauchy).

Therefore, $$a,b \in \mathbb{R} \Rightarrow (a+b)\in \mathbb{R}$$

Closure of Multiplication:

Similarly, for $(a_n),(b_n)$ being Cauchy sequences representing nonzero $a,b$, one would need to show that $(a_n b_n)$ is Cauchy. This can be done similarly to the first, just a bit more finicky.

So, as before, we have, since $(a_n),(b_n)$ are Cauchy, for some $N_1, N_2 \in \mathbb{N}$

• $|a_m - a_n| < \varepsilon \;\;\; \forall \varepsilon > 0, \forall m,n > N_1$
• $|b_p - b_q| < \varepsilon \;\;\; \forall \varepsilon > 0, \forall p,q > N_2$

We let $N = \max\{N_1,N_2\}$. Since Cauchy sequences are bounded, we'll also take $|a_n|, |b_n| \leq M$ for some positive $M$. (We particularly want nonzero for our proof, but $|a_n|,|b_n| \geq 0$ by definition, so "positive" also works.)

As before, too, we'll take the above inequalities that come from $(a_n),(b_n)$ being Cauchy with $\varepsilon/2M$ as opposed to just $\varepsilon$ for convenience.

Then, as we want $(a_nb_n)$ to be Cauchy, we consider and note:

$$|a_m b_m - a_n b_n| = |a_m b_m - a_n b_m + a_n b_m - a_n b_n |$$

Invoking the triangle inequality,

$$|a_m b_m - a_n b_m + a_n b_m - a_n b_n | \leq |a_m b_m - a_n b_m | + | a_n b_m - a_n b_n |$$

We do some factoring and note:

$$|a_m b_m - a_n b_m | + | a_n b_m - a_n b_n | = |b_m| \cdot |a_m - a_n | + |a_n| \cdot |b_m - b_n|$$

Since $(a_n),(b_n)$ are Cauchy, we take them as being less than $\varepsilon/2M$ for $m,n>N$. Since they're also bounded, $|a_n|,|b_m| \leq M$. Thus,

$$|b_m| \cdot |a_m - a_n | + |a_n| \cdot |b_m - b_n| < |b_m| \cdot \frac{\varepsilon}{2M} + |a_n| \cdot \frac{\varepsilon}{2M} = \frac{\varepsilon}{2M} (|b_m| + |a_n|) \leq \frac{\varepsilon}{2M} (M+M) = \varepsilon$$

Thus, for all $\varepsilon > 0$ and $m,n > N$,

$$|a_m b_m - a_n b_n| < \varepsilon$$

on the premise $(a_n),(b_n)$ are. Thus, $(a_n b_n)$ is Cauchy. Since $a,b \in \mathbb{R} \setminus \{0\}$ implies $(a_n b_n)$ is Cauchy, then, as $(a_n b_n)$ represents $a\cdot b$, we thus have

$$a,b \in \mathbb{R} \setminus \{0\} \Rightarrow (a\cdot b) \in \mathbb{R} \setminus \{0\}$$

Well, it is not so complicated if you grant that $\mathbb Q$ is closed under addition and multiplication. Since $\mathbb Q$ is dense in $\mathbb R$, the uniformly continuous functions $\mathbb Q \times \mathbb Q \rightarrow \mathbb Q$ of addition and multiplication extend uniquely to functions $\mathbb R \times \mathbb R \rightarrow \mathbb R$ on the completion.

The general principle behind this is that if $X$ is a dense subset of a metric space $\widetilde{X}$, and $f$ is a uniformly continuous map of $X$ into a complete metric space $Y$, then $f$ extends uniquely to a continuous map from $\widetilde{X}$ to $Y$.

This is how you define addition and multiplication in $\mathbb R$, so you immediately get that $\mathbb R$ is closed under these operations.

It's also by continuity that you can check that the same laws that hold for $\mathbb Q$ (associativity, commutativity, distributivity) remain true for $\mathbb R$.