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.
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 $\alpha$.