How to derive Church-Kleene ordinal Crossing-out: (How does one prove the existence of Church-Kleene ordinal? Also, why is it labeled as $\omega_1^{CK}$?
And why is it first ordinal not hyperarithmetical, and is the first admissible ordinal after $\omega$?)
Edit: OK, I will cross out the first question and the second question. And I will change my question to this:
How does one prove that $L_{\omega_1^{CK}}$ is the model of Kripke-Platek set theory, and how does one prove that it is first admissible ordinal after $\omega$? (So why can't $\omega+1$ can be a model of KP set theory?) And why is it first ordinal not to be hyperarithmetical?
And somehow extraneous question: Why are admissible sets labeled "admissible"? Curiosity.
 A: Let me elaborate on the last paragraph of Boris' answer above.
Define $\omega_1^{CK}$ (as usual) as the least noncomputable ordinal - that is, the least ordinal $\alpha$ such that no computable well-ordering of the natural numbers has ordertype $\alpha$. You ask:

Why is $\omega_1^{CK}$ the first non-hyperarithmetic ordinal?

Well, certainly every computable ordinal is hyperarithmetic, so this really boils down to: why is $\omega_1^{CK}$ non-hyperarithmetic (precisely: why is there no hyperarithmetic relation on the natural numbers with ordertype $\omega_1^{CK}$)? This is an important theorem of Spector: in general, a structure can have a hyperarithmetic copy without having a computable copy, but for ordinals we have "hyperarithmetic = computable" (Incidentally, this result was later strengthened by Montalban).
There are a couple different ways to prove Spector's theorem. Sacks' book has what I think is the standard proof. A different proof (which I cooked up myself, but I think is probably folklore) is to show how, given a hyperarithmetic (even $\Sigma^1_1$) ordinal $\alpha$, there is a computable ordinal $\beta$ with $\beta>\alpha$; this then implies that $\alpha$ itself has a computable copy (just "chop off" a computable copy of $\beta$ at the appropriate point). Constructing such a $\beta$ takes a bit of an argument, but it's self-contained and doesn't invoke Kleene notations.

Specifically, here's how we construct such a $\beta$. (I'm assuming here that we've already proved "hyperarithmetic = $\Delta^1_1$" - if not, see Sacks' book. It's worth noting that this too can be proved without invoking notations at all, if we define "hyperarithmetic" as "computable from the jump sequence along some computable well-ordering.") Suppose $\alpha$ is a hyperarithmetic ordinal. Fix a $\Sigma^1_1$ formula $\varphi(x, y)$ defining a well-ordering $\mathcal{W}$ on a ($\Sigma^1_1$) subset of $\omega$ with ordertype $\alpha$. By the normal form theorem for $\Sigma^1_1$ sentences, $\varphi(x, y)$ can be assumed to have the form "The tree $S_{x, y}$ has an infinite path" for some uniformly computable family of trees $(S_{a, b})_{a, b\in\mathbb{N}}$ on $\omega$. (WLOG suppose $\alpha$ is infinite.)
Associated to $\varphi$ is a tree of approximations, $T_\varphi$. A node $\sigma$ on $T_\varphi$ is a triple $(m, F, (p_{(a, b)})_{(a, b)\in F})$ consisting of:


*

*A natural number, $m$ (the "length" of the node).

*A finite set $F$ of pairs of natural numbers, intended to represent part of $\mathcal{W}$.

*For each pair $(a,b)\in F$, a finite path $p_{(a, b)}$ of length $m$ in the tree $S_{a, b}$. (Remember, this tree has an infinite path iff $a<_\mathcal{W}b$.)
The ordering on $T_\varphi$ is defined by setting $$(m, F, (p_{(a, b)})_{(a, b)\in F})<(n, G, (q_{(a, b)})_{(a, b)\in G})$$ iff 


*

*$m<n$,

*$F\subseteq G$, and

*For each $(a, b)\in F$, we have $p_{(a, b)}\prec q_{(a, b)}$.
By the usual coding shenanigans, this tree can be thought of as a computable subtree of $\omega^{<\omega}$.
This tree is not well-founded: in particular, it has infinite paths corresponding to the actual structure of $\mathcal{W}$ (and sub-relations). However, attached to each node $\sigma\in T_\varphi$ is a finite linear order - namely, the "$F$" part. These "amalgamate" as we move up the tree (this is the requirement $F\subsetneq G$ in the definition of the ordering above), and - since any infinite path corresponds to a subrelation of $\mathcal{W}$ - we have that the direct limit along any path is well-founded. 
Form a tree of descending sequences through that direct limit (it's a bit messy to write this down, but it's fine); this tree is computable and well-founded. Taking its Kleene-Brouwer ordering gives a computable well-ordering of ordertype $\ge\alpha$.
(This argument can be generalized to prove a stronger result, which was what I actually needed when I cooked it up (Theorem 1.4).)

Arno Pauly has also given (corollary 81) a proof of Spector's theorem using the machinery of admissible representations, but I haven't read it in detail yet so I don't know how different it is.
A: An ordinal $\alpha$ is admissible if $L_\alpha\models\Sigma_0$-collection. Here, for any set theory $T$, $M\models T$ means that for any set $S$, if $S\in M$ then all $S'\in M$ for all $\phi(S)\vdash S'$ if $\phi$ is a formula on sets that $T$ either assumes or proves to exist (the formula not the sets; trying to avoid double meanings). $\Sigma_0$-collection is an assumption that if $\phi$ is a formula with parameters in $N$ for some $N$, then there exists a set $X:=\{S|\phi(S)\}$.
$\omega$ is the first admissible ordinal because $L_\omega$ is the set of all finite sets in the constructible hierarchy, where "finite" means that it has finite length and all elements of it are also finite. Clearly if $S_{1,...,n}$ is a sequence of finite sets, then $\{S_1,...,S_n\}$ is also finite, therefore $L_\omega$ is the minimal model which satisfies "if $S_{1,...,n}$ are in $L_\omega$, then $\{S_1,...,S_n\}$ is also in $L_\omega$".
The same cannot be said about $L_{\omega+1}$ because, among other things, $L_\omega\in L_{\omega+1}$ and $L_\omega$ is an infinite set, so $L_{\omega+1}$ doesn't satisfy finite set grouping axiom, and neither does it satisfy any other form of collection.
$\omega^{CK}_1$ is the least ordinal above $\omega$ such that $L_{\omega^{CK}_1}\models\Sigma_0$-collection, since for any computable/recursive well-order $\langle\mathbb{X},\prec\rangle$ on sets $X\subseteq\mathbb{X}$, if $\mathbb{X}\in L_{\omega^{CK}_1}$ then $\text{field}(\prec)\in L_{\omega^{CK}_1}$. The same follows for all $\omega^{CK}_{\alpha+1}$ except that we have well-orders restricted to $\langle\mathbb{X},\prec\rangle\cong\langle\beta,<\rangle$ for some $\omega^{CK}_\alpha\leq\beta\in\omega^{CK}_{\alpha+1}$, which is why admissibles are ususally not limits of admissibles, aka in the form $\omega^{CK}_\alpha$ for a limit $\alpha$. I am saying "usually", because the ones that are admissibles as well as limits of admissibles are called recursively inaccessible ordinals.
$\omega^{CK}_1$ is the first non-hyperarithmetical ordinal, because it is the least unrecursive well-ordering of $\omega$. The recursive well orderings are those of the form of $\Delta^1_1$-defineable (aka hyperarithmetic) subsets of $\omega$.
