Possible not countable extension of the natural numbers? This question comes from:Is $1234567891011121314151617181920212223......$ an integer?
We define $\mathcal{A}$ as the set of infinite strings of digits
$$
\bar a_i=a_0 a_1a_2a_3\cdots a_i \cdots  \qquad a_i \in \{0,1,2,3,4,5,6,7,8,9\}
$$
We can easily show, with Cantor's diagonal proof, that $\mathcal{A}$ is a not countable set. 
We define :
$$
\bar a_i =\bar b_i \iff a_i =b_i \quad \forall i
$$ 
and a function $|\cdot| : \mathcal{A} \rightarrow \mathbb{N}\cup \{+\infty\}$
$$
|\bar a_i|=\sum_{i=0}^\infty{10^{\,i}\,a_i}
$$
The subset of  $\mathcal{A}$:
$$
\mathcal{N}=\{\bar a_i : |\bar a_i| \in \mathbb{N}\}=\{\bar a_i : \exists n \quad \forall i>n \quad  a_i=0\}
$$ 
consists of finite strings of not null digits. Ignoring the (infinite) null digits at the right side we can define on it the usual operations of sum and product and so it is a ring isomorphic to $\mathbb{N}$.
We can extend these operations to all elements of $\mathcal{A}$ using a recursive definition: for the sum $\bar a_i+\bar b_i=\bar c_i$ with:
$$
q_0=(a_0+b_0) \div 10 \qquad; \qquad c_0=(a_0+b_0)\mod 10
$$
$$
q_i=(a_i+b_i+q_{i-1}) \div 10  \qquad; \qquad c_i=(a_i+b_i+q_{i-1})\mod 10
$$
and for the product with:
$$
\bar a_i \cdot 0=0 \qquad \bar a_i \cdot (\bar b_i +1)=\bar a_i \cdot  \bar b_i +\bar a_i
$$

(Here there is some abuse of notation as noted by @anorton, but I think that the meaning is obvious).
If we define the successor function as $ s(\bar a_i) = (\bar a_i)+ 1$ (with $1=1000\cdots)$ we see, as noted in the answer of  @Asaf, that $(\bar 9)=(999 \cdots)$ has not a successor in the usual sense. We can solve this problem excluding $(\bar 9)$ from $\mathcal{A}$, or defining $s(\bar 9)=(\bar 9)$. I don't know if this second possibility is compatible with PA, but it seems to me that the Peano's 6th axiom does not request $s(n)\ne n \quad \forall n$.

We can define an order relation in $\mathcal{A}$ that coincides, in $\mathcal{N}$, with the order on natural numbers, defining:
$$
\bar a_i < \bar b_i \iff \exists \bar c_i \quad s.t. \quad \bar b_i=\bar a_i + \bar c_i
$$
or
$$
\bar a_i < \bar b_i \iff \exists n \quad s.t. \quad  (b_i\le  a_i \quad \forall i \ge n) \; \land \; (\bar a_i \ne \bar b_i)
$$

As noted in the answer of  @Asaf, this second definition is not a total order on $\mathcal{A}/\mathcal{N}$. Again I don't know if this is compatible with PA, but, if not, we can use the first definition.

The question is: we can consider the set $\mathcal{A}$ (or $\mathcal{A}/\{(\bar 9)\}$) so equipped as a non countable model for the Peano's axioms of the natural numbers?
 A: What would be the successor of the $999\ldots$? The language of $\sf PA$ include (and in some sense, it is based on) a successor function, which is equivalent to adding $1$.
What would $999\ldots+1$ be?
Additionally, your second definition of $<$ fails. Since $989898\ldots$ and $999\ldots$ are of course comparable, and $989898\ldots\leq 9999$, but there is no $n$ after which the digits of $989898\ldots$ are strictly smaller than the digits of $999\ldots$.
A: Your definition of addition makes excellent sense if you write the numbers as infinite to the left,
$$ \ldots a_4 a_3 a_2 a_1 a_0 $$
Your recursive definition of multiplication is not complete; it only fixes the result of multiplying something with a finite number. But instead one can extend the pencil-and-paper multiplication algorithm for base-ten numbers to your infinite strings of digit in essentially the same way as you treated addition.
If you do that, you will have reinvented the 10-adic integers. That's a perfectly good uncountable ring that extends integers.
However, it is not a model of Peano Arithmetic, since by your definition of addition we have
$$ \ldots 99999 + \ldots 00001 = \ldots 00000 $$
so $0$ is a successor of $\ldots 99999$, but PA explicitly requires that $0$ isn't a successor of anything.
The 10-adics also contain some, but not all, proper fractions, such as
$$ \frac 13 = \ldots 66667 $$
since if you add three of those you get $\ldots 00001$ and keep carrying 2 into infinity.
A: The Asaf's answer proof that the proposed set is not an extension of real numbers because if we exclude $(\bar 9)$ we have the same problem for $(899999 \cdots)$ and so forth.
On the other hand, if we pose $s(\bar 9)=(\bar 9)$ we have $s(89999\cdots)=s(\bar 9)$ that contradict the Axiom 8.
