Is true that $\bigcup α=α$ if $\alpha$ is a limit ordinal? 
Definition
Given a set $A$ the membership relation on $A$ is the relation defined by the identity
$$
\in_A:=\{a\in A\times A:a_1\in a_2\}
$$
Definition
A set $A$ is said transitive if ech its element $a$ is there contained, that is
$$
\forall a\in A:a\subseteq A
$$
Definition
A set $\alpha$ is an ordinal number if it is a transitive set and if it is well ordered by the membership relation.

So as here shown, If $X$ is a set whose elements are transitive sets then $\bigcup X$ is transitive too. Moreover if the elements of $X$ are ordinals then the elements of $\bigcup X$ are ordinals too because if $y$ is an element of $\bigcup X$ then there exists $Y\in X$ such that $y\in  Y$ but $Y$ is an ordinal and any element of an ordinal is an ordinal too so that $y$ is an ordinal and in particular the membership relation is a well order for $X$ because if $\alpha$ and $\beta$ are two distinct ordinals the one (only one) of the inequality
$$
\alpha<\beta\,\,\,\text{or}\,\,\,\beta<\alpha
$$
holds. Therefore we conclude that the union $\bigcup X$ of a set of ordinal $X$ is an ordinal.
Now if $\alpha$ and $\beta$ are two ordinals such that $\alpha$ is a proper subset of $\beta$, that is
$$
\alpha\subset\beta
$$
then it is possible to show that $\alpha\in\beta$ and so $\alpha <\beta$.
Moreover by the axiom of union any element $x$ of a set $X$ is contained in the union of $X$, that is
$$
\forall x\in X:x\subseteq\bigcup X
$$
So we conclude that if $X$ is a set of ordinals then
$$\alpha\le\bigcup X
$$
for any $\alpha\in X$: indeed, if $\alpha\in X$ then $\alpha\subseteq\bigcup X$ and so if $\alpha\subset\bigcup X$ then (remember that $\bigcup X$ is an ordinal and that any proper subset of an ordinal is an ordinal there contained) $\alpha< X$ whereas $\alpha=X$ trivially. Otherwise if $\beta$ is an ordinal greater than the elements of $X$ then $\alpha\in\beta$ for any $\alpha\in X$ and thus by transitivity $\alpha\subseteq \beta$ which implies that
$$
\bigcup X\subseteq \beta
$$
so that by the previous argumens we conclude that $\bigcup X\le\beta$.
So finally we concldue that if $X$ is a set of ordinals then $\bigcup X$ is the smallest ordinal greater than the elements of $X$.
However in this Von Eitzen's answer is proved that if $\alpha$ is the successor of an ordinal $\beta$ then $\bigcup\alpha$ is equal to this $\beta$ and so it is less then $\alpha$ so that I argue that the $\bigcup X$ is not the least ordinal greater than or equal to all elements of $X$ but this is stated in the text Introduction to Set Theory by Karel Hrbacek and Thomas Jech which prove it with the argumentations I gave above: here you can see this by your self.
So where's the wrong? Moreover observing that
$$
\bigcup \omega=\omega
$$
I ask also if any limit ordinal (an ordinal is said limit ordinal if it is not the successor of any ordinal) is equal to its union. So could someone help me?
 A: So above it is proved that for any ordinal number $\alpha$ the set $\bigcup\alpha$ is the least ordinal greater than any element of $\alpha$ and thus if $\alpha$ is obviously greater than any its element then necessarily the inequality
$$
\bigcup\alpha\le\alpha
$$
must holds. Now if $\alpha$ is the successor of any ordinal is not hard to show that it has a maximum element so that this element must be exactly equal to $\bigcup\alpha$ if this is the least ordinal greater than any element of $\alpha$ so that
$$
\alpha=\bigcup\alpha+1
$$
because for any ordinals $\eta$ it cannot exist an ordinal $\theta$ such that
$$
\theta<\eta<\theta+1
$$
and thus if the last equality did not holds then it wolud be
$$
\bigcup\alpha+1<\alpha
$$
which means that $\bigcup\alpha$ is not the least ordinal greater than any element of $\alpha$. On the contrary if $\alpha$ is a limit ordinal then it is possible to show that it has not a maximum element so that it must be
$$
\bigcup\alpha=\alpha
$$
because it is not possible that the inequality
$$
\bigcup\alpha<\alpha<\bigcup\alpha+1
$$
holds and so if the last equality did not hold then it would be
$$
\bigcup\alpha+1<\alpha
$$
which means that $\bigcup\alpha$ is not the least ordinal greater than $\alpha$.
