What are the sets $S_n=\omega-n$ called?

Question

What are the sets $S_n$ where $S_n:=\omega-n$ called?

I explain better: if ordinals are defined in this way

$0=\varnothing$

$1=\{\varnothing\}=\{0\}$

$2=\{0,1\}$

$n=\{0,1,..,n-1\}$

$\omega=\{0,1,2,..\}$

1. What are the sets $S_n=\omega-n=\{n,n+1,n+2,...\}$ called?

2. For every $S_n$, $|S_n|=\aleph_0$? Why can't there exist an $S_n$ with a finite number of elements (I can think about it in my mind but I don't see a way to find which are those elements)?

3. Does $S_{\omega}=\varnothing$? (Intuitively I would say yes, because since the elements of $n\cup S_n$ are the elements of $\omega$, $\omega -\omega=\varnothing$ because if I take from the naturals all the naturals I have the empty set.)

Maybe what I said has no meaning, someone can explain better this, what I don't see.

Answer 1

1. The sequence $(1, 2, \ldots, n - 1)$ is an initial segment of the sequence $(1, 2, 3, \ldots)$, and $(n, n+ 1, n + 1 \ldots)$ is an end-segment. I suppose you could call the set $\{n, n+ 1, n + 1 \ldots\}$ an end-segment too, taking the usual order as understood.
2. Why can't there exist a $S_n = \{n, n +1, n + 2 \ldots\}$ with a finite number of elements? Because there is an obvious bijection between the elements of $S_n$ and $\mathbb{N}$ (map $n + k \in S_n$ to $k \in \mathbb{N}$): so, by definition, $S_n$ is countably infinite.
3. What is $S_{\omega}$? You have only defined $S_n$ for finite $n$. If the idea is that $S_{\omega} =_{\mathrm{def}} \omega - \omega$ and the 'minus' still indicates set difference, then yes, $S_{\omega}$ is empty.

Answer 2

1. I would probably call them final segments of $\omega$, but I don't think I've seen these sets referred to as anything in particular before.
2. Since $S_n \subseteq \omega$ then $S_n$ is either finite or countably infinite. If it were finite, then $\omega$ would be the union of two finite sets, which is absurd!
3. If you define $S_\omega$ to be $\omega \setminus \omega$ (in analogy to $S_n$), then, yes, $S_\omega = \emptyset$. If you define $S_\omega$ to be $\bigcap_{n < \omega} S_n$, then again $S_\omega = \emptyset$, since $n \notin S_{n+1} \supseteq S_\omega$ for all $n \in \omega$.

Answer 3

I am not awareness a particular name. As a whole, this collection is the set of end segments/tails of $\omega$.

If I had to give these sets a name, I woe probably say that $S_n$ is the "$n$th tail of $\omega$".

These are all infinite because otherwise $n\cup S_n$ is the union of two finite sets - impossible.