Checking a set with the Axiom of Integers

Question

In learning the axiom of integers which are the following:

1. Z has a binary operation +
2. Z had a binary operation *
3. Distributive law - a * (b + c) = a * b + a * c
4. There is a subset Z^+ called the positive integers satisfying the following a). Z^+ is closed under addition b). Z^+ is closed under multiplication c). For every a in Z, exactly one of the following holds a is in Z^+, -a is in Z^+ or a = 0
5. Well-ordering Axiom

Using these axioms,there was a problem from my textbook I found to be tricky, which is the following:

Let X be the set with only one element, 0. We provide it with two binary operations: 1). 0 + 0 = 0 2). 0 * 0 = 0

Let X^+ be the empty set. Which axioms fail?

So going through all the axioms

1). Since we are given the binary operation 0 + 0 = 0, it does have an identity of 0, is associative, commutative, and the last part of Axiom 1

2). Since we are given the binary operation 0 * 0 = 0, this one has me very confused as Axiom 2 states that we should get an identity of 1, but if X only has 0, there is no identity that is equal to 1?

3). I think that it does follow the distributive property 0 * (0 + 0) = 0 * 0 = 0

4). I think this axiom is fine for X^+ being closed under addition and multiplication, but it fails for part (c) where for every z in Z, exactly one of the following holds. As the empty set is the set containing no elements, none of the three cases hold

5). The Well ordering principle states that every non-empty set of the positive integers contains a least element. Would this axiom also fail as our original set containing only 0 which you can consider does have a least element or due to our subset being the empty set that this axiom fails?

Any help is greatly appreciated!

Thanks for reading!

Answer 1

Actually they satisfy all the axioms you've stated.

Since we are given the binary operation 0 * 0 = 0, this one has me very confused as Axiom 2 states that we should get an identity of 1, but if X only has 0, there is no identity that is equal to 1?

Firstly, none of your axioms mention any identities. So the question is moot. But even if there are also axioms stating that there is an additive identity and a multiplicative identity, it does not mean that they must be different, unless you have an axiom that says that. In particular, in the structure you're given $0$ is the multiplicative identity for $X$.

it fails for part (c) where for every z in Z, exactly one of the following holds. As the empty set is the set containing no elements, none of the three cases hold.

No. (c) refers to $Z$, not $Z^+$.

5). The Well ordering principle states that every non-empty set of the positive integers contains a least element. Would this axiom also fail as our original set containing only 0 which you can consider does have a least element or due to our subset being the empty set that this axiom fails?

To even state "well-ordering" we need to have the ordering on the structure. To say that $X^+$ is well-ordered under $<$ has nothing to do with integers. Don't get confused; positive integers are well-ordered under the standard ordering. If you have another structure like $X^+$ and you want to say that it is "well-ordered", well, under what?

We can assume that $X$ has the implicit ordering, since there is only one possible ordering. What is it? In that case we need to determine whether any non-empty subset of $X^+$ has a least element. Think of quantifiers using game semantics; for me to prove it, you move first and give me any non-empty subset of $X^+$, then I move next and prove that it has a least element. Whoever can't make a move loses. So if there is no non-empty subset of $X^+$, you lose.