Smallest next real number after an integer This might be a silly question, but is it possible at all for n.00000...[infinite zeros]...1 to be the next real number after n? If not, why not?
Firstly, I know (I think) that $$\lim_{x\to \infty} \frac{1}{10^x} = 0$$ but I'm not talking about taking its limit. Surely I'm not required to.
The obvious rebuttal is that n.00000...[infinite zeros]...1 (let's just call this number c) is that c divided by two will be between n and c, which is obviously a closer number to n than c. But why is c necessarily divisible? There are special exceptions for other numbers. For instance, if b=0 in $\frac ab$, we say it's undefined.  Why can't $\frac c2$ be similarly undefined? Or something similar to the notion that infinity divided by two is still infinity?
OR, does c equal n.0, similar to how 0.999... = 1?
Apologies if this question has been asked a million times before (I was not able to find it asked quite this way) or if you find it stupid.

You can define a number like this one if you consider hyperreal numbers: and it would be written $n + \epsilon$, which is greater than n but less than any other number greater than n.
Indeed, if you could define n.00000...[infinite zeros]...1 , then you should be able to define n.00000...[infinite zeros]...2 and many other numbers. This 0.000[infinite zeros]1 is called $\epsilon$ and is the first infinitesimal part. It can be multiplied by any real and would still be infinitesimal. $\epsilon^2$ is an even smaller infinitesimal part than the first one.
But because of the hierarchy, you would also have $n < ... < n + \epsilon^3 < n + \epsilon^2 < n + \epsilon$ and even with ordinals you could not get around this problem.
So even in this theory you cannot find "the" one after n, only one in a subhierarchy.
