Is the limit of a sequence of ordinals distinct from the ordinal of the union of nested sets up to the limit?
Consider ordered, nested subsets $g_n\subsetneq g_{n+1}$ of integers whose order types obey the relations:
$\text{otp}(g_0)=1$ and
$\text{otp}(g_{n+1})=\omega^{\text{otp}(g_{n})}$
Therefore the order type of $\lim_{n\to\infty}\text{otp}(g_n)$ is $\epsilon_0$
Is it correct / the same thing to say:
$\displaystyle\text{otp}\left(\bigcup_{n\in\Bbb{N}}g_n\right)=\epsilon_0$?
In other words, is the order type of the infinite union of these sets the same as the limit of the sequence of order types of these sets? It seems there should be a slight distinction between the order type and its limit. It seems like I'm asking whether or not $\epsilon_0$ is an open set. Intuition tells me it will be an open set so I need to carefully observe the distinction above.
Actually, if I understand your question correctly, the answer is no. Consider the subsets $g_n\subseteq\varepsilon_0+\varepsilon_0$ defined as follows: $$g_n=[0,\omega\uparrow\uparrow(n-1))\cup[\varepsilon_0,\varepsilon_0+\omega\uparrow\uparrow n)$$ (here $\alpha\uparrow\uparrow n$ is $\alpha^{\alpha^{\dots^\alpha}}$ with the tower of height $n$ (this is called the up-arrow notation) and $[\alpha,\beta)$ is the half-open interval). Then $g_n$ clearly has order type $\omega\uparrow\uparrow(n-1)+\omega\uparrow\uparrow n=\omega\uparrow\uparrow n$, as you request, but $\bigcup_{n\in\mathbb N}g_n=\varepsilon_0+\varepsilon_0$ has order type higher than $\varepsilon_0$.
The reason why such a construction works is because we are adding elements out of order - your claim would be true if we required that $g_{n+1}$ is a so-called end-extension of $g_n$, which means that every element of $g_{n+1}\setminus g_n$ is greater than every element of $g_n$.