I recently asked this question for which I got the reply that $$x = 1 + 1 + 1 \cdots \textrm{ till } \infty$$ is not a real number. Now I want to ask what $x$ exactly is? If $x = \infty$, then does $\infty \in \mathbb{R}$? I don't think that $x \in \mathbb{C}$ as complex numbers can be expresssed in the form $a + \mathrm{i}b$ where $a, b \in \mathbb{R}$ and $\mathrm{i} = \sqrt{-1}$.
Also, why are normal arithmetic operations not valid for $x$?
the sum $\sum_0^{\infty}1$ diverges to infinity. The normal arithmetic operations are defined on $\mathbb{R}$, but since $x$ isn't in $\mathbb{R}$, the operations are not defined for it. This is for the same reason you can't make arithmetical operation on $x=\text{cat}$; The operations are simply not defined.
Note that you can define some operations on $\mathbb{R} \cup \{\infty\}$ in a way that makes sense: For example, you can define
$$\forall x \in \mathbb{R}, x + \infty = \infty + x = \infty$$
$$\infty + \infty = \infty$$
These operations preserve the associativity and commutativity of addition in $\mathbb{R} \cup \{\infty\}$. However, the operations with infinity must be defined beforehand, and you can't assume they are just there.
Also note that in math it is possible to define things however you like, but most definitions aren't "interesting" (e.g. if associativity is not preserved)
The $\sum\limits_{n \geq 0}1$ diverges. And $\infty$ is not a number, but a concept. The normal arithmetic operations are not valid for $x:=\sum\limits_{n \geq 0}1$, because it's divergent. As in the linked question, when you take $x+1=x$ and subtract $x$, you are basically doing $\infty - \infty$, which is not valid.
By definition, $x=\sum_{k\geq 1} 1=\lim_{n\to\infty}n$. So the existence of $x$ as a real number relies on the sequence $(n)_{n\in\mathbb{N}}$ to converge in $\mathbb{R}$, which it doesn't. You could however compactify $\mathbb{R}$ to $\hat{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$ to force this sequence to converge. Think of the compactification of $\mathbb{R}$ as a circle as opposed to a straight line. Then the limit exists, but you lose the validity of your arithmetic operations on the way since the compactified real numbers are no longer a field. Also the limit is the added point $\infty$ which is not in $\mathbb{R}$ by construction of $\hat{\mathbb{R}}$.
With "compactify $\mathbb{R}$" above I mean the one-point compactification of $\mathbb{R}$, which is indeed homeomorphic to the circle $S^1$.
