Problem $2.18$, Rudin's RCA - Painfully Set Theoretic 
Problem $2.18$: This exercise requires more set-theoretic skill than the preceding ones. Let $X$ be a well-ordered uncountable set which has a last element $\omega_1$ such that every predecessor of $\omega_1$ has at most countably many predecessors. ("Construction": Take any well-ordered set which has elements with uncountably many predecessors, and let $\omega_1$ be the first of these; $\omega_1$ is called the first uncountable ordinal.) For $\alpha\in X$, let $P_\alpha[S_\alpha]$ be the set of all predecessors (successors) of $\alpha$, and call a subset of $X$ open if it is a $P_\alpha$ or an $S_\beta$ or a $P_\alpha \cap S_\beta$ or a union of such sets. Prove that $X$ is then a compact Hausdorff space. (Hint: No well-ordered set contains an infinite decreasing sequence.)

*

*Prove that the complement of the point $\omega_1$ is an open set which is not $\sigma$-compact.

*Prove that to every $f \in C(X)$ there corresponds an $\alpha\ne\omega_1$ such that $f$ is constant on $S_\alpha$.

*Prove that the intersection of every countable collection $\{K_n\}$ of uncountable compact subsets of $X$ is uncountable. (Hint: Consider limits of increasing countable sequences in $X$ which intersect each $K_n$ in infinitely many points.)

Let $\mathfrak M$ be the collection of all $E \subset X$ such that either $E \cup \{\omega_1\}$ or $E^c \cup \{\omega_1\}$ contains an uncountable compact set; in the first case, define $\lambda(E) = 1$; in the second case, define $\lambda(E) = O$. Prove that $\mathfrak M$ is a $\sigma$-algebra which contains all Borel sets in $X$, that, $\lambda$ is a measure on $\mathfrak M$ which is not regular (every neighborhood of $\omega_1$ has measure $1$), and that $$f(\omega_1) = \int_X f\ d\lambda$$
for every $f \in C(X)$. Describe the regular $\mu$ which Theorem 2.14 associates with this linear functional.


The problem certainly demands more set-theoretic skill than what I possess. My goal is to solve the problem in full, and that would start by understanding what it's saying. I request you to kindly help me with explanations/hints, and I will keep updating this post with my progress on this problem as I'm able to understand and do more parts of it.

Thoughts: First, they take $X$ to be a well-ordered (every non-empty subset has a least element) uncountable set with last element $\omega_1$. What do they mean by last? What is meant by "let $\omega_1$ be the first of these"?
They have defined exactly the open sets (i.e. the topology) in $X$. We must prove that $X$ is compact and Hausdorff. Hausdorff-ness is clear from Oliver's answer. For compactness, note that the topology on $X$ is the order topology, and by Theorem 27.1 of Munkres, we are done.
Update:
For (1): Note that $P_{\omega_1} = X\setminus \{\omega_1\}$, since $\omega_1$ is the last element. So the complement of $\omega_1$ is open. We want to show that it is not $\sigma$-compact. How do I do that?
For (2): Suppose $f\in C(X)$, i.e. $f$ is continuous on $X$. I'm not sure what the codomain of the function is, so I'm assuming $\mathbb C$. It looks like we want to find $z\in\mathbb C$ so that $f^{-1}(\{z\})$ is an $S_\alpha$ set for $\alpha < \omega_1$.
For (3): I found this answer, which seems to work.
Lastly, I could use hints on how to go about the final part of the problem about measures.
Thanks!
 A: The juicy parts (the compactness of $X$ and the analysis on $X$ are left for you since you are very enthusiastic on seeing this problem though and may need help only with a few details here and there.


*$X=[0,\omega_1]$ with the order topology is a Compact and Hausdorff space:

a. Compactness if $X$: If you are not well versed in this business of orders and ordinals, a quick read though Point Set Topology books will help you. There are many classic books. Here is just a reference if you need to freshen up on order relations, the week order principle and the order topology: Munkres, J. R. Topology, 2ns ed., pages 24-28 (definitions and examples); pages 63-66 (well order principle, what it is and there stuff); pages 84-86 (order topology); and pages 172-174 (proof of compactness of intervals of the form $[a,b]$ in linearly ordered spaces with the supremum property, of which your space $X$ and also the real line are examples). All this will give you  more than enough about the topology of $X=[0,\omega_1]=P_{\omega_1}\cup\{\omega_1\}$.
b. $X=[0,\omega_1]$ is Hausdorff: suppose $x,y\in X$ and $x\leq y$, w.l.o.g assume $x<y$. If $y$ is a successor of $x$ then $[0,y)=P_y$ is an open neighborhood of $x$ and $(x,\omega_1]=S_x$ is an open neighborhood of $y$; moreover, $[0,y)\cap(x,\omega_1]=\{z\in X: x<z<y\}=\emptyset\}$ (because $y$ is a successor or $x$). If $y$ is not a successor of $x$, then there is $z\in X$ between $x$ and $y$,that is, there is $z\in X$ such that $x<z<y$. Then $[0,z)$ is and open neighborhood of $x$, and $(z,\omega_1]$ is an open neighborhood of $y$ and $[0,z)\cap(z,\omega_1]=\emptyset$.


*

*$[0,\omega_1)$ is not $\sigma$-compact:

Since open sets of the form $(x,y)=[0,y)\cap(x,\omega_1]$, and of the form $[0,y)$,  with $0\leq x<y<\omega_1$, are countable (why?) all compact subsets of $[0,\omega_1)$ are countable (being covered by finite union of such sets). The union of coubtable sets, being countable, can't cover $[0,\omega_1)$ which is uncountable by the choice of $\omega_1$.
Obs:
c. A consequence of (1) is that for any compact set $K\subset[0,\omega_1]$, $K$ is   uncountable iff $\omega_1\in K$ and $\omega_1$ is a cluster point of $K$, i.e. for any $\alpha<\omega_1$, $(\alpha,\omega_1)\cap K\neq\emptyset$. Indeed, if $\omega_1$ is an isolated point of $K$, then $K\setminus\{\omega_1\}$ is  a compact set contained in $[0,\omega_1)$ and so, $K=(K\setminus\{\omega_1\})\cup\{\omega_1\}$ is countable. Conversely, if $K$ is uncountable, then $K$ is not contained in any set of the form $[0,\alpha]$, $\alpha<\omega_1$; hence $K\cap(\alpha,\omega_1)\neq\emptyset$.
d. Incidentally, the observation above also shows that any increasing sequence $A=\{a_n:n\in\mathbb{N}\}\subset[0,\omega_1)$ converges to a point $\beta\in[0,\omega_1)$. Indeed, $A$ is bounded above by $\omega_1$. Let $b$ be the first element of $\{\beta\in X:a_n\leq \beta,\,n\in\mathbb{N}\}$. Then $a_n\xrightarrow{n\rightarrow\infty}b$ and so, $\{b,a_n:n\in\mathbb{N}\}$ is countable and compact. Hence $b\neq\omega_1$.



*The intersection of a countable collection of uncountable compact sets is uncountable:

Suppose $\mathcal{K}=\{K_n:n\in\mathbb{N}\}$ is a sequence of uncountable compact sets in $X$. Observation (c) above implies that $\omega_1$ is a cluster point of each $K_n$. Consider first $K_1\cap K_2$. Let $\alpha_0<\omega_1$. Then there is $\alpha_1\in K_1$ such that $\alpha_0<\alpha_1<\omega_1$. In turn, this implies that there is $\alpha_2\in K_2$ such that $a_1<\alpha_2<\omega_1$. Proceeding by induction, we obtain a sequence $A=\{\alpha_n:n\in\mathbb{N}\}$ such that $\alpha_n<\alpha_{n+1}$, $\alpha_n\in K_1$ for $n\equiv1\operatorname{mod}2$, and $\alpha_n\in K_2$ for $n\equiv0\operatorname{mod}2$. Observation (d) implies that $A$ converges to some $\beta\in[0,\omega_1)$. Clearly $\beta$ is a cluster point of both $K_1$ and $K_2$; hence, $\beta\in K_1\cap K_2$ ( $K_j$, $j=1,2$ is closed). Define $K'_n=\bigcap^n_{\ell=1}K_\ell$. Then, each  $K'_n$ is compact and uncountable, and  since $\bigcap^\infty_{n=1}K'_n=\bigcap^\infty_{n=1}K_n$,  we may assume without loss of generality that the sequence $\mathcal{K}$ is monotone non increasing, which we do now. Arguing as before, we get that for any $\alpha\in[0,\omega_1)$, we find a sequence increasing sequence $\{a_n:n\in\mathbb{N}\}$ such that $\alpha<a_n\in K_n\cap[0,\omega_1)$. Then, we have that $b=\sup_na_n\in (\alpha,\omega_1)$ is a cluster point each $K_n$; hence $b\in K=\bigcap_nK_n$. This shows that $\omega_1$ is a cluster point of $K$, and by (c), $K$ is uncountable.



*For every $f\in\mathcal{C}(X)$, there is $\beta_f\in[0,\omega_1)$ such that $f$ is a constant on $(\beta,\omega_1]$:

For any $n\in\mathbb{N}$,     the set $U_n=\big\{\alpha\in X:|f(\alpha)-f(\omega_1)|<\frac1n\big\}$ is open and contains $\omega_1$. Let $\alpha_1\in U_1\setminus\{\omega_1\}$. Once $\alpha_1<\ldots<\alpha_n<\omega_1$ have been constructed, choose $\alpha_{n+1}\in U_n\cap(\alpha_n,\omega_1)$ (this is possible since any open set containing $\omega_1$ contains an interval of the form $(\alpha,\omega_1)$. By (d), $\beta=\sup_n\alpha_n\in[0,\omega_1)$. It follows that for all $\beta<\gamma\leq\omega_1$, $f(\gamma)=f(\omega_1)$, for $\gamma\in\bigcap_nU_n$.

All this take care of the topological aspects of the problem.


*I leave to the OP to show that the family
$$\mathcal{F}=\{E\subset X: \text{either}\,E\cup\{\omega_1\}\,\text{or}\, E^c\cup\{\omega_1\}\,\text{contains an uncountable compact set}\}$$
is a $\sigma$-algebra containing the open sets.
Hint:


*

*Clearly $\mathcal{F}$ is closed under complementation, and $X\in\mathcal{F}$.

*(3) and observation (c) will be useful to show that $\mathcal{F}$ is closed under countable intersections.

*It will be useful to check that open sets in $X$ are countable union of sets of the form $(x,y)$, $[0,y)$ and $(x,\omega_1]$, where $0\leq x<y\leq \omega_1$.



*I also leave the details that $\lambda$ is indeed a measure on $\mathscr{F}$.

Once the OP proves this, it is easy to check that  that for any $0\leq a<\omega_1$, we have $\lambda\big((a,\omega_1)\big)=1$, $\lambda(\{\omega_1\})=0$,  and $\lambda([0,a])=0$.
From this, and part (2) of the OP, it follows that for any $f\in\mathcal{C}(X)$,
$$\int_X f\,d\lambda=\int_{(\alpha_f,\omega_1]}f\,d\lambda=f(\omega_1)\lambda\big((\alpha_f,\omega_1]\big)=f(\omega_1)$$
The lack of regularity of $\lambda$ is also easy to check as $\lambda(\{\omega_1\})=0$ and $\lambda(U)=1$ for any open neighborhood $U$ of $\omega_1$.

The regular measure $\mu$ obtained from application of Riesz-Markov representation theorem to the linear map $f\mapsto f(\omega_1)$ satisfies the property $\mu(A)=\mathbb{1}_{A}(\omega_1)$ for every Borel set $A$ (this is the $\delta$-measure supported by $\{\omega_1\}$).
A: For Hausdorff, you can use that $X$ is linearly ordered. So if $\alpha<\beta$, use a combination of predecessor/successor open sets to separate them. Keep in mind that $\alpha$ has an immediate successor but $\beta$ might not have an immediate predecessor.
For compactness, assume the open sets are $P$'s or $S$'s since they form a basis. Then it is helpful to notice that the two types of open sets are nested: if $\alpha\leq\beta$, then $P_\alpha\subseteq P_\beta$ and $S_\beta\subseteq S_\alpha$. Then split the open cover into the two types and think about how much they cover. The well-foundedness of the ordering will help with the $S_\alpha$.
As an aside, if you are still thinking about what the ordering looks like, it is like an uncountable version of the ordering of the natural numbers 0,1,2,...$\omega$ where you put an infinite number $\omega$ on the end which is larger than all of the finite numbers. So here the open sets $P_n$ are $0,1,2,...n$ and the open sets $S_n$ are $n,n+1,n+2,....\omega$, and you get the whole space or just $\omega$ if you pick $\omega$ instead of a finite number.
