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.)

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    $\begingroup$ Are you familiar with the formal definition(s) of $\mathbb{R}$ (Cauchy sequences, Dedekind cuts, ...)? This is a great question, and part of the answer is that we have to go down to precisely such a formal definition. $\endgroup$ Dec 12, 2018 at 2:40
  • $\begingroup$ I'm somewhat familiar with Cauchy sequences. Dedekind cuts, not so much, sadly. $\endgroup$ Dec 12, 2018 at 2:42
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    $\begingroup$ "Another possibility is to start from some rigorous axiomatization of Euclidean geometry (Hilbert, Tarski, etc.) and then define the real number system geometrically. From the structuralist point of view all these constructions are on equal footing." (From wiki) $\endgroup$ Dec 12, 2018 at 2:44
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    $\begingroup$ That's no problem, the proofs are easier with Cauchy sequences anyway: Suppose $a=(a_1,a_2,a_3,\dots)$ and $b=(b_1,b_2,b_3,\dots)$ are real numbers, i.e. Cauchy sequences of rationals. Define $a+b$ to be the sequence $(a_1+b_1,a_2+b_2,a_3+b_3,\dots)$. Prove that it is Cauchy, therefore also a real number. $\endgroup$
    – user856
    Dec 12, 2018 at 3:21
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    $\begingroup$ Define a real number as an equivalence class of Cauchy sequences of rationals, where two sequences are equivalent are equivalent if they converge to the same value. Define addition and multiplication elementwise. Show that the result is Cauchy and does not depend on which representatives of the equivalence classes are chosen. $\endgroup$
    – saulspatz
    Dec 12, 2018 at 3:26

5 Answers 5


As I seem to be so fond of saying, it all comes down to definitions. Specifically, before you can answer the question "Why are the real numbers closed under addition and multiplication?", one needs to answer the question "What is a real number?" There are lots of ways of doing this–I'll outline two common approaches.

Axiomatic Approaches

The real numbers can be defined axiomatically, i.e. we simply list the properties that we want the real numbers to have, then work with those properties directly. The "usual" axiomatization is to define the real numbers to be the (unique, up to isomorphism) totally ordered, Dedekind-complete field. Basically, if we let $\mathbb{R}$ denote the set of real numbers, then $\mathbb{R}$ is defined by the properties that:

  1. $\mathbb{R}$ is a field; that is, there are two binary operations $+$ and $\cdot$ defined on $\mathbb{R}$ which obey the usual field axioms;
  2. $\mathbb{R}$ is totally ordered; that is, there is a relation $\le$ defined on $\mathbb{R}$ allowing us to compare any two real numbers and (importantly) this order is compatible with the field operations in the sense that
    • $x \le y \implies x+z \le y+z$, and
    • $0\le x,y \implies 0 \le xy$.
  3. $\mathbb{R}$ is Dedekind-complete; that is, every nonempty bounded set in $\mathbb{R}$ has a least upper bound.

If $\mathbb{R}$ is defined in this manner, then the answer to your question is trivial: $\mathbb{R}$ is a field because we have defined it to be so, and fields are closed under addition and multiplication. These properties are "baked in" to the definition of the real numbers.

It is worth noting that other axiomatizations are possible. For example, Tarski's axiomitization, which defines $\mathbb{R}$ as a linearly ordered abelian group with some other nice properties. Under this axiomatization, $\mathbb{R}$ is, by definition, closed under addition, but no multiplication operation is defined a priori. I am not as familiar with this construction, but the above cited Wikipedia article suggests that Tarski was able to define a multiplication operation and show that it behaved as expected, making $\mathbb{R}$ into a field. I suspect that one could also go the route of defining the group $\mathbb{R}$ in terms of the Tarski axioms, then obtaining the field $\mathbb{R}$ as the field of fractions.

A More Constructive Approach

(In the interest of brevity, the following discussion is very loosey-goosey. It can mostly be made rigorous, but I'm not sure that it would be terribly useful to do so.)

The downside of a axiomatic approach is that you are kind of asserting the existence of of a mathematical object which has the properties that you want. It might be more satisfying to build the real numbers from more basic building blocks. We still need some axioms as a starting point, be we can start from a much more primitive place, say the Peano axioms or the ZFC axioms for set theory.

From the Peano axioms, we get the natural numbers immediately, so let's start there. From this point of view, $\mathbb{N}$ is a fairly simple object: about the only thing we get from the definition is a successor operation and zero, which gives us an order on $\mathbb{N}$, but not a lot else. However, from these, we can build up $\mathbb{N}$ as a monoid (there is an additive structure with identity, but no inverses).

From the natural numbers, we can define the integers: first, we define an equivalence relation on $\mathbb{N}\times \mathbb{N}$ by declaring that $(a,b)\sim (c,d)$ if $a+d = c+b$. The integers are then elements of the set $\mathbb{N}\times\mathbb{N}\,/\,{\sim}$, i.e. equivalence classes modulo this relation. The basic idea is that $(a,b)$ represents the number $a-b$ (so $(0,b)$ is the integer $-b$, for example). Since subtraction and negative numbers are not defined by default, everything gets done in terms of addition. When the dust settles, we get a group $(\mathbb{Z},+)$. This group has all the nice additive properties which we expect from the integers, so we can now forget that the integers are "really" equivalence classes of ordered pairs of natural numbers. Note that $\mathbb{N}$ embeds nicely into $\mathbb{Z}$, and that the order on $\mathbb{N}$ can be extended to $\mathbb{Z}$. So, now we have an ordered group. Hoo-ray!

On the integers, it is now possible to define a multiplication operation which is compatible with the addition operation and the order relation. For example, we can define multiplication here as repeated addition, then use some fancy induction. The problem with the integers is that most integers don't have multiplicative inverses, so we need to patch that up. We do this in much the same way that we got the integers. This time, we define an equivalence relation on pairs of integers by declaring that $(p,q)\sim (r,s)$ if $ps = qr$, then mod out by this relation. This gives us the set $\mathbb{Q} := \mathbb{Z}\times\mathbb{Z}\,/\,{\sim}$, i.e. the field of fractions of $\mathbb{Z}$. Again, $\mathbb{Z}$ can be embedded into the rationals in a way that preserves all of the operations as well as the order.

Finally, to get the real numbers, we want some kind of notion of completeness. The usual constructions are via Dedekind cuts or Cauchy sequences. At the end of the day, the two constructions can be shown to be equivalent, but, as a guy who does analysis on metric spaces, I am more comfortable with the construction via Cauchy sequences.

Since this is the heart of your question, I'll be a little more detailed here.

Let $|\cdot| : \mathbb{Q} \to \mathbb{Q}_{\ge 0}$ denote the usual absolute value on $\mathbb{Q}$. What we would really like to do is use the absolute value to define a metric on $\mathbb{Q}$, but we don't have the real numbers in hand yet, so this would be a little circular. However, the absolute value does define a valuation on $\mathbb{Q}$, which is morally the same thing as a norm which can induce the moral equivalent of a metric.

Definitions: Let $(a_n)_{n=1}^{\infty}$ be a sequence of rational numbers.

  • We say that $(a_n)$ is Cauchy if for any rational number $\varepsilon > 0$, there exists a natural number $N$ such that $m,n \ge N$ implies that $|a_m - a_n| < \varepsilon$.
  • We say that $(a_n)$ converges to a limit $L\in\mathbb{Q}$ if for any rational $\varepsilon> 0$, there exists some $N$ such that $n\ge N$ implies that $|a_n-L|<\varepsilon$.

We would like to live a world where Cauchy sequences converge to limits. However, there are examples of Cauchy sequences in $\mathbb{Q}$ which do not converge in $\mathbb{Q}$. Consider, for example, the sequence of partial sums $$ S_N := \sum_{n=0}^{N} \frac{1}{n!}, $$ which is Cauchy but has no rational limit. To rectify this situation, we once again build an equivalence relation.

Definition: We say that a sequence $(a_n)$ is a null sequence if $a_n$ converges to $0$. That is, for any rational $\varepsilon > 0$, there exists some $N$ such that $n \ge N$ implies that $|a_n| < \varepsilon$.

Now we declare an equivalence relation on the set of sequences. Specifically, we say that $(a_n) \sim (b_n)$ if the sequence $(a_n - b_n)$ is a null sequence. It is a little bit of work to show that this is an equivalence relation, but it works out. Hoo-ray! We can now define $\mathbb{R}$ to be $\mathscr{C}(\mathbb{Q})\,/\,{\sim}$, where $\mathscr{C}(\mathbb{Q})$ denotes the set of Cauchy sequences.

A real number is then an equivalence class of Cauchy sequences. To finish the construction, we need an order relation, as well as addition and multiplication operations. Since your question is about the additive and multiplicative closure of $\mathbb{R}$, I'll ignore the order relation and note only that addition and multiplication are defined pointwise, i.e.

Definition: Let $a = [(a_n)]$ and $b = [(b_n)]$ be two equivalence classes of Cauchy sequences represented by $(a_n)$ and $(b_n)$, respectively. Then define

  • $a+b := [(a_n + b_n)]$, and
  • $a\cdot b := [(a_n \cdot b_n)]$.

To show that the reals are closed, it is necessary to show that $(a_n + b_n)$ and $(a_n \cdot b_n)$ are both Cauchy. This isn't too bad to do, so I'll leave it as an exercise. Once the exercise is completed, the question is answered. :)


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.

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    $\begingroup$ Where do your $\varepsilon$s live? If you are not careful, your construction runs the risk of being circular. Typically, when we talk about metric spaces, the target space of the metric is the real numbers, which you haven't defined yet. $\endgroup$
    – Xander Henderson
    Dec 12, 2018 at 4:08
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    $\begingroup$ @XanderHenderson: fair point, I should have written it. When defining Cauchy/convergent sequences, the $\varepsilon$s live in the original ordered, Archimedean set. $\endgroup$ Dec 12, 2018 at 4:09

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.

  • $\begingroup$ I believe that the most of the answers thus far involving Cauchy sequences was because of a comment I made in the question's comments that I'm more familiar with them than Dedekind cuts. Not that I don't appreciate the alternate viewpoint though! ^_^ $\endgroup$ Dec 13, 2018 at 7:01
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    $\begingroup$ @EeveeTrainer: Another way that I've seen proposed on Math SE before to simplify the Dedekind cut construction is to use the route $\mathbb{N} \to \mathbb{Q}_{\ge0} \to \mathbb{R}_{\ge0} \to \mathbb{R}$, which avoids a lot of the case-splitting due to sign. I still prefer the Cauchy sequence approach, because it is in some sense more constructive. $\endgroup$
    – user21820
    Dec 13, 2018 at 7:34
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    $\begingroup$ @user21820: True that you avoid the splitting that way; that said, I was following Dedekind’s exposition. The disadvantage I see with Cauchy sequences is that you have to mod out by an equivalence relation, and then prove that the operations are well-defined, which you do not have to do with cuts if you specify which side will have the endpoint in the case of a rational cut. Not sure in what sense cauchy sequences are “more constructive”, though. If anything, the Dedekind cuts construction actively reifies the points on the line. $\endgroup$ Dec 13, 2018 at 7:52
  • $\begingroup$ @ArturoMagidin: To reduce the difficulty with modding out the equivalence, we can simply construct all Cauchy sequences and define arithmetic and equivalence on them and prove that they form a field (with equivalence in place of equality), and then it is logically obvious that we can collapse the equivalence classes if we wish. The constructiveness also arises in this way; computable reals can be defined as computable Cauchy sequences, and arithmetic on them is computable (excluding division by zero), unlike when defined as computable Dedekind lower cuts, where even addition is uncomputable. $\endgroup$
    – user21820
    Dec 13, 2018 at 8:04
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    $\begingroup$ @user21820: I am aware of that; you still end up having to establish that the quotient “is” what you want it to be. In short, it’s really a matter of picking your poison”. $\endgroup$ Dec 13, 2018 at 14:42

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

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

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    $\begingroup$ Your answer is basically correct, and I laud your effort to take on the challenge of the exercise I left. The only tweak that I might make is to note that it doesn't exactly make sense to say that a Cauchy sequence $(a_n)$ converges to a real number $a$. The Cauchy sequence is (in a meaningful, though somewhat sloppy sense) the real number $a$. It would be better to say that $a$ is a real number represented by a given Cauchy sequence, and that $b$ is a real number represented by a different Cauchy sequence. Otherwise, what you have written looks good to me (after cursory review). (+1) $\endgroup$
    – Xander Henderson
    Dec 13, 2018 at 3:28

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 $\mathbb R$.

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    $\begingroup$ Your answer is misleading. You cannot talk about density in the reals without having already defined the reals, which is the underlying issue under inquiry. Also, without the reals you cannot define "metric space". There is a weaker definition of "metric space" that permits completion of rationals via Cauchy sequences, but one must be careful. $\endgroup$
    – user21820
    Dec 13, 2018 at 7:36
  • $\begingroup$ Okay sure, I didn't write this answer to give a comprehensive answer with all details written. Just to give an explanation to OP that once you made sense out of what $\mathbb R$ is as a metric space, the addition and multiplication operations on $\mathbb R$ are nothing mysterious and naturally arise from that on $\mathbb Q$. $\endgroup$
    – D_S
    Dec 13, 2018 at 7:48
  • $\begingroup$ Yes, it's a good thing to remember that density of rationals in reals implies a certain rigidity where continuous functions on reals are determined by their values on rationals. Of course, that's after constructing the reals, which is a more fundamental problem. Anyway thanks for responding! =) $\endgroup$
    – user21820
    Dec 13, 2018 at 7:53

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