Can a Peano Set have two or more zeros? I repeat the Peano Axioms:


*

*Zero is a number.

*If a is a number, the successor of a is a number.

*zero is not the successor of a number.

*Two numbers of which the successors are equal are themselves equal.

*If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.
Suppose to have two isomorph "copies" of the natural numbers $\mathbb{N}':=\{0',1',2'...\}$ and $\mathbb{N}'':=\{0'',1'',2''...\}$.
Then the set $NUMBERS:=\mathbb{N}'\cup \mathbb{N}''$ with "Zero"$:=0'$ and the "natural" successor for each element in any of the two sets, seems to satisfy the axioms.
Yes, P5 is very strange now because, it says that when I start with a set which I know to contain at least $0'$ and every successor of the numbers in it, automatically contains $0''$ which is not a successor of any number.
If this way of reasoning is allowed we could also use a number of copies of $\mathbb{N}$ indicized by a continuous index so there will be TWO NOT ISOMORPHIC Peano sets.
Because this sounds very strange to me, It's possible that there is a problem in my argument.  What do you think?
 A: Note that from the fifth axiom it follows that if $x\neq 0$ then $x=S(y)$ for some $y$. 
Therefore if $x,y$ are both $0$'s (and are distinct) then at least one of them is a successor. Which means that one of them is not a $0$.

I think that the source of the confusion here is the fact that $0$ is seen as just any number. The truth is that $0$ is a constant in the formal language of arithmetic. And in a given structure there can only be one interpretation of a constant. Not two or more.
A: In logic, $0$ is a constant, not a property. That is, we can say "$x=0$" rather than "$x$ is a zero."
Even in second order logic, the statement of induction is $0\in S$. Not, "for all zeros $z, z\in S$."
You can, of course, interpret lots of language in lots of different ways. But mathematics uses "$0$ is a natural number" in a very specific way, and if you want to interpret that phrase differently, you are free to do so, but you are likely to confuse mathematicians and fail to communicate with them unless you are very explicit.
A: I'm not fully sure I understand your question, but perhaps you will find this argument enlightening: consider the set
$$\mathbb{N}' = \{0', s(0'), s(s(0')), \ldots\}$$
which contains $0'$ and all its successors. (This is the same $\mathbb{N}'$ you defined in the question.) Hopefully it is clear that $\mathbb{N}'$ contains the successor of every element in $\mathbb{N}'$:
$$\forall n\in\mathbb{N}'\ s(n)\in\mathbb{N}'$$
So axiom 5 applies to $\mathbb{N}'$, which tells us that $\mathbb{N}'$ contains all numbers, or in other words, axiom 5 tells us that if $n$ is a number, $n\in\mathbb{N}'$. That in turn means anything not in $\mathbb{N}'$ is not a number.
Now, if you want, you can postulate the existence of another object, like $0''$, and even a set $\mathbb{N}''$ of that object and its successors. But you cannot claim that $0''$ (or any of its successors) is a number (as defined in the Peano axioms) without contradicting the result of the preceding argument, and thus contradicting axiom 5.
You can construct sets which contain numbers and other things, like $\mathbb{N}'\cup\mathbb{N}''$. This set contains every number, because every number is in $\mathbb{N}'$ and by the definition of the union operation, anything in $\mathbb{N}'$ is in the union of $\mathbb{N}'$ and any other set. It also contains a bunch of other things which are not numbers, namely the objects in the set $\mathbb{N}''$. Nothing says that every object in the set $S$ referenced in axiom 5 must be a number.
A: Theorem
Suppose we have zeroes $0$ and $0'$ in a Peano set such that:


*

*$0$ is a number.

*$0'$ is a number

*If x is a number, the successor of x is a number.

*$0$ is not the successor of any number.

*$0'$ is not the successor of any number.

*Two numbers of which the successors are equal are themselves equal.

*If a set S of numbers contains $0$ and also the successor of every number in S, then every number is in S.

*If a set S of numbers contains $0'$ and also the successor of every number in S, then every number is in S.
Therefore $0=0'$, and there is a unique zero in every Peano set (defined in the obvious way).
Proof
We can easily prove by induction (7) that all non-$0$ numbers have a predecessor.
Suppose $0\neq 0'$. Therefore $0'$ must have a predecessor. But this contradicts (5). Therefore, we must have $0=0'$. 
