I can't wrap my head around Peano's 5'th axiom I'm currently reading about Peano's axiom and the construction of the natural numbers. The fifth Axiom of Peano does not make sense to me:

*

*If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ is also in $T$, then $T$ is the set of Natural numbers $\mathbb{N}$.

I know that this axiom exists in order to exclude elements which could have the following property:
$$s(a) = b$$ $$s(b) = a$$
These elements would not violate axioms $1$ to $4$ however they don't have the desired "natural number" properties hence axiom $5$ has been created to allegedly clean these elements up. I also know that axiom $5$ is supposed to work as follows:

*

*If $0$ is in our set $T$ then $s(0)$ is also in $T$ but if $s(0)$ is in $T$ then $s(s(0))$ is also in $T$ and so on. Thus we declare this set to be the natural numbers.

Now to my actual question: suppose we throw in $a$ and $b$ and $0$ into $T$. Since we do not yet know what the natural numbers are, we can't exclude $a,b$ either (if my thinking is correct). Then for all $n$ in $T$ the succesor $s(n)$ is also in $T$. As previously stated $s(0), s(s(0)), ...$ are all in in $T$ (the actual natural numbers) but if such elements $a$ and $b$ are also in T then their successor $s(a)$ and $s(b)$ are also in $T$ which is  a true statement since $s(a) = b$ and $s(b) =a$.
Axiom $5$ should exclude such elements but the way I am stating it, it seems like these elements do not violate axiom $5$.
 A: Axiom 5, in conjunction with the other four axioms, tells you that:

*

*Every natural number other than $0$ is a successor; and

*You can “reach” every natural number by starting with $0$ and applying the successor function finitely many times.

To see that the five axioms prevent you from having “cycles” with the successor function, note first that we cannot have $s(a)=a$ for any $a$. To see this, consider
$$T = \{n\in\mathbb{N}\mid n\neq S(n)\}.$$
By Peano’s third axiom, $0\in T$ (since $0$ is not a successor, so $0\neq S(0))$.
Assume $k\in T$. Then $k\neq S(k)$. By Peano’s fourth axioms, this implies $S(k)\neq S(S(k))$; hence $S(k)\in T$.
By Peano’s Fifth Axiom, $T=\mathbb{N}$.  In particular, $S(a)\neq a$ for all $a\in\mathbb{N}$.
Similarly, if $S(a)=b$ and $S(b)=a$, then $S(S(a))=a$. Consider
$$T=\{a\in\mathbb{N}\mid a\neq S(S(a))\}.$$
Again, $0\in T$, since $0$ is not a successor. Assume $k\in T$. Then $k\neq S(S(k))$, hence $S(k)\neq S(S(S(k))$, so $S(k)\in T$. Thus, $T=\mathbb{N}$, and you cannot have the two elements you propose.
You can similarly verify that there cannot exist elements $a_1,\ldots,a_m$ with $S(a_i) = a_{i+1}$ for $i=1,\ldots,m-1$, and $S(a_m)=a_1$.
A: (Posted after previous posting was accepted.)

I know that this axiom exists in order to exclude elements which could
have the following property:
$~~~~s(a) = b$
$~~~~s(b) = a$

$\cdots$

Axiom $5$ should exclude such elements but the way I am stating it, it
seems like these elements do not violate axiom $5$.

In this case, $a, b\neq 0$ and $\{a, b\}$ is an isolated, closed loop under the successor function $s$, i.e. neither of the elements of  $\{a, b\}$ are accessible from any other natural numbers. This contradicts the fact that induction holds on $N$.
It can be shown that induction holds on any set $X$ with $x_0$ being the "first" element under the successor function $f$ if and only if there exists no such isolated (unconnected?) subsets $Y\subset X$ under $f$, with $x_0 \notin Y$. (See my formal proof using a form of natural deduction here, 228 lines) .
