Problems with proof of $\omega_1=\bigcup\{X_\xi|\xi\in\omega_1\}$ I have the proof of following lemma, which I do not really understand:
Let $(X,<)$ be a well orderd uncountable set.
Let $\omega_1=\{\xi\in X|X_\xi\,\,\text{countable}\}$ and $X_\xi=\{\alpha\in X|\alpha <\xi\}$
Lemma:

$\omega_1=\bigcup\{X_\xi|\xi\in\omega_1\}$

My main issue seems to be, that the elements of $\omega_1$ are elements of $X$ such that $X_\xi$ are countable, while the elemens of $\bigcup\{X_\xi|\xi\in\omega_1\}$ are countable sets. And I do not understand why we can 'compare' these sets in the first place. It somehow seems self-referential.
Following proof is given:
Is $\alpha <\xi\in\omega_1$, we have $X_\alpha\subseteq X_\xi$. So $X_\alpha$ is countable and it is $\alpha\in\omega_1$. So we have $X_\alpha\subseteq\omega_1$.
Which proofs $\omega_1\supseteq\bigcup\{X_\xi|\xi\in\omega_1\}$.
And I do not understand this conclusion. 
I do not understand why both sets have compareable (the same) elements...
For the other part:
Is $\xi\in\omega_1$ then is $X_\xi$ countable. Then is $\xi^+\in\omega$ since $X_{\xi^+}=X_\xi\cup\{\xi\}$ is countable and $\xi\in X_{\xi^+}$. Hence $\subseteq$.
I understand every detail of both proofs, but I do not understand why it shows what we desire...
Can you give an explanation?
Thanks in advance.
 A: I'm not sure where the confusion is, since you say you understand every detail, but as you don't understand the whole, I suspect you might not understand why we do every detail. Therefore I will try to rephrase the proofs with a little more words, which will hopefully make the purpose more clear.
First let's check that both $\omega_1$ and $\bigcup\{X_\xi\mid \xi\in\omega_1\}$ are subsets of $X$:
A set $a$ is an element of $\omega_1$ if $a\in X$ and the set $\{\xi\in X\mid \xi<a\}$ is countable.
A set $b$ is an element of $\bigcup\{X_\xi\mid\xi\in\omega_1\}$ if $b\in X_\xi$ for some $\xi\in\omega_1$, so then $b\in X$, since $X_\xi\subset X$.
So they are "comparable" in that they are both subsets of $X$.
To show that they are the same subset, we show that $\omega_1\supseteq \bigcup\{X_\xi\mid\xi\in\omega_1\}$ and $\omega_1\subseteq \bigcup\{X_\xi\mid\xi\in\omega_1\}$ both hold. This implies that the sets are equal by the Axiom of Extensionality.
For $\supseteq$, let $\alpha\in\bigcup\{X_\xi\mid\xi\in\omega_1\}$, then $\alpha\in X_\xi$ for some $\xi\in\omega_1$. We have to show that $\alpha\in\omega_1$. Since $\alpha\in X_\xi$, by definition of $X_\xi$ we have $\alpha<\xi$. If for some $\beta\in X$ it holds that $\beta\in X_\alpha$, then $\beta<\alpha<\xi$, and by transitivity of $<$ it then follows that $\beta<\xi$, and thus $\beta\in X_\alpha$. In other words $X_\alpha\subseteq X_\xi$. Since subsets of countable sets are countable, this means that $\alpha\in\omega_1$ by definition of $\omega_1$, as was needed.
For $\subseteq$, let $\alpha\in\omega_1$, then we need to show that $\alpha\in\bigcup \{X_\xi\mid \xi\in\omega_1\}$. This means finding a $\xi\in\omega_1$ such that $\alpha\in X_\xi$. We know $X_{\alpha^+}$ contains $\alpha$, since $\alpha<\alpha^+$. Furthermore, since $\alpha\in\omega_1$, we know that $X_\alpha$ is countable. But then $X_{\alpha^+}=X_\alpha\cup\{\alpha\}$ is countable, and thus $\alpha^+\in \omega_1$. So then $X_{\alpha^+}$ is one of the sets in our union and contains $\alpha$.
