Are these axioms of real number strict？

Question

After comparing with some other textbooks about introductory real analysis, I find that many books' content about axioms of real numbers are not strict (at least for me, I think they are not strict). For example, in ross's $\textit{Elementary Analysis}$, the axioms he states are as follow:

and in bruckner's $\textit{Elementary Real Analysis}$, they state like this:

In both case, I find that they do not mention that $1$ should be assumed to not equal to $0$. My point is that we from these axioms, we can not argue that $1\neq0$, and also from these axioms, we can only conclude that there exists maybe two elements, $0$ and $1$. However, we can see that if we assume $0=1$, the set $R= \left\{ 0 \right\}$ with two operation $+: R \times R\to R,0+0=0$, $*:R \times R\to R, 0*0=0:$can also satisfy the field axioms, and furthermore, it can satisfy the order axioms.

So, I think they are not strict. Can anyone tell me if my arguments are right?

Addition: Thanks for @Henry remind me that in Ross's book, he adds in the late that the R should have at least two elements,so my construction doesn't work for this example. However, I don't know how to prove that $0 \neq 1$.

Answer 1

This is an odd question for a simple reason: the set of axioms you've listed are a set of axioms that $\mathbb{R}$ satisfies, but they are not a set of defining axioms. Neither book you quote from pretends that these are a complete set of axioms defining $\mathbb{R}$.

Referring to Ross's book (which I have a copy of, and thus I can actually refer to) for example, you've copied exactly the set of axioms defining a (possibly trivial) field. In the next paragraph, Ross continues to define a field as a set with at least 2 elements satisfying the properties A1-A4, M1-M4, and DL. In the following paragraph, Ross gives an additional set of axioms that every ordered field must satisfy. Over the next couple of chapters, he describes how to construct the real numbers from the rational numbers (with an eye towards completeness and Dedekind cuts).

From this perspective, it should not be surprising that choosing a small subset of the properties that $\mathbb{R}$ satisfies doesn't uniquely distinguish $\mathbb{R}$. It might be an interesting exercise to try to come up with lots of structured sets that satisfy these axioms.

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Let us now consider the question at the heart of your post. I rephrase it here.

Suppose a set with two binary operations $+$ and $\cdot$ satisfies the axioms A1-A4, M1-M4, and DL. Do these axioms on their own force $0 \neq 1$?

No, these axioms do not force $0 \neq 1$. The set consisting of the element $e$, defined with $e \cdot e = e$ and $e + e = e$ satisfies all of these axioms. You've noted this in your question.

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Suppose that we include the next paragraph of Ross's book, which requires that a field have at least $2$ elements.

Suppose a set $S$ with the binary operations $+$ and $\cdot$ satisfies the axioms A1-A4, M1-M4, DL, and that in addition $S$ contains at least two elements. Does this force $0 \neq 1$?

Now the answer is yes. To see this, suppose that $0 = 1$. As $S$ contains at least $2$ elements, there is some other element $a \neq 0$.

Then we must have both that $a \cdot 1 = a$ and $a \cdot 0 = 0$, but as $0 = 1$ this forces $a = 0$. This is a contradiction.

(In this, I have assumed that $a \cdot 0 = 0$. You can prove this from the field axioms, and this is done in Theorem 3.1 of Ross for example).

Thus $1 \neq 0$.

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This leads to a natural question.

Suppose that a set $S$ has at least two elements and satisfies the axioms A1-A4, M1-M4, and DL. Does this force $S$ to be the real numbers?

No! The set of rational numbers $\mathbb{Q}$ satisfies these properties. (Indeed, this is the center of the narrative in Ross at this point). But there are lots of other structures too.

For example, consider $S$ to be the numbers $0, 1, 2$, and consider the binary operator $+$ to be addition modulo $3$ the binary operator $\cdot$ to be multiplication modulo $3$. This is a set of $3$ numbers, and this is also a field. The non-obvious thing here is that $2^{-1} = 2$, as $2 \cdot 2 \equiv 1 \bmod 3$.

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This line of thinking, where you try to determine a set of axioms that specifies $\mathbb{R}$, is a very healthy line of thinking. This is not so trivial!