Edit : Ok, "is not a real number" is a very different statement from "does not exist.". My initial thought is "does not exist.". Sorry, I never know about something like the hyperreal number before.
From the definition of the ordinal number ω, ω is the smallest ordinal number that larger than all finite ordinal number. That mean all finite ordinal number < ω.
My assumption : Then I think about that "Is there a smallest ordinal number A that $\frac{1}{n}>\frac{1}{A}$ for all finite ordinal number n ?" The $\frac{1}{n}>\frac{1}{A}$ mean $A>n$. That mean A is ω. But does this mean $\frac{1}{ω}$ exist ?
However, when I watch the video about the Dedekind Cuts by Elliot Nicholson on Youtube (Part 1,Part 2,Part 3,Part 4), there are always rational numbers between any 2 different real numbers regardless if those two number is rational or irrational.
If the number $\frac{1}{ω}$ is exist in real number line, there will be no rational number between 0 and $\frac{1}{ω}$ or between $1-\frac{1}{ω}$ and 1 or etc. which is contradict to the Dedekind Cuts.
So, I think the number $\frac{1}{ω}$ doesn't exist in real number line and my assumption is wrong but I don't understand where and how my assumption is wrong. Could you explain me ?
Edit : I am sorry that I ask with my limited knowledge. But if I don't ask, I will not know more. My initial thought is that "Is there a smallest number A (in whatever number system) that $\frac{1}{n}>\frac{1}{A}$ for all positive integer n ?" but I am not sure if I should ask exactly like my initial thought or not. That is why I ask like that.
Ordinal numbers are not real numbers. Ordinal arithmetic is different from the basic arithmetic we see with the natural numbers, and while it does extend it, it extends it in a different way.
If you think about the natural numbers, then $\frac 1n$ does not make any sense anyway, since it is not a natural number. We need to add these objects to our world, but before we can do that, we need to verify that $\frac1n$ is going to be unique in a meaningful way. In other words, we want that $\frac1nn=1$, or more generally, if $\frac1nnk=k$. This follows from the fact that $nm=nk\implies m=k$, namely that multiplication is cancellative on the natural numbers.
Ordinal arithmetic, however, does not have this property, and it's not commutative either. Case in point, $\omega=2\cdot\omega=3\cdot\omega=n\cdot\omega$ for any $0<n<\omega$, but we don't know that $1=2=3=\dots=n=\dots$, that's just false.
We sometimes talk about the surreal numbers, or other extended number systems which extend the real numbers into which we can embed $\omega$ in a semi-meaningful way. But there are two problems here:
The embedding is respecting ordinal arithmetic. Specifically, these "extended number systems" are often forming a field, or at the very least a commutative ring, but the ordinal arithmetic is not commutative as we saw. While there is a notion of "Hessenberg arithmetic" (sometimes called "natural arithmetic") which is commutative which does embed into these number systems (most notably the surreal numbers), it does not reflect the natural way in which we think about adding and multiplying linear orders, which ultimately is what being an ordinal number is all about.
Even if we can embed $\omega$ into such a field, it is not a "convex" embedding. There is no tight upper bound to the natural numbers. In any type of field. To see that, simply note that if $F$ is an ordered field and $x=\sup\Bbb N$, then $x-1<n$ for some $n$, and therefore $x<n+1$. So we effectively pick a point which is more (or sometimes less) arbitrary which lies above all the repeated sums of $1$ with itself, and call that $\omega$. So in what sense is that point going to be $\omega$ the ordinal? It's not. It's as representative of it as $1$ would be if we consider the embedding $n\mapsto 1-\frac1{2^n}$. And in this case, $1$ represents $\omega$, so $\frac1\omega=\frac11=1=\omega$. And we can see that this is not very helpful at all.
The thing to remember here is that numbers are usually some sort of mathematical abstraction of quantity. But not all quantities are comparable. We don't often find ourselves asking "what's larger, 1km or 1kg?", and so different abstractions will behave differently and may or may not be comparable with one another.