A quick question. Is $$\textrm{ot}(\bigcup\limits_{\gamma <\lambda}\alpha_{\gamma})=\bigcup\limits_{\gamma <\lambda}\textrm{ot}(\alpha_{\gamma})?$$ where $\textrm{ot}$ stands for the order type (and $\lambda$ can be limit ordinal or not).
It seems like a nice property which can be very much false, but I don't know neither how to prove it nor can I find a counterexample.
... Something funny has just happened to me: I believed for many years that, if $a$ and $b$ are subsets of a well-ordered set $x$, then $$\textrm{type}(a \cup b) \leqslant \textrm{type(a)} + \textrm{type}(b)$$
However, this is not correct (and I have should know better $\ldots$). I was talking to a student, trying to prove the above inequality, and something was not right ("what happens if the set $a \cup b$ has a maximum, and such maximum is an element of $a$ ?"). Indeed: the true statement is $$\textrm{type}(a \cup b) \leqslant \textrm{type(a)} + \textrm{type}(b) + 1$$
To check that the first inequality is incorrect, you may consider $x = \omega.2 + 1$, $a = \omega \cup \{\omega.2\}$ and $b = x \setminus a$.
(To prove the correct inequality, one has just to proceed with induction on the type of $a \cup b$, this isn't difficult.)
Edited: The second inequality is also incorrect !!! See comments below of @asatzhh.