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I am a beginner of ordinals and I don't know any powerful techniques in it. I come across with a problem about the first uncountable ordinal like this.

Let $X$ be a set of uncountable cardinality. Using the Principle of Well Order we have a well ordering $\le$ on X(and $<$ means $\le$ but not euqual). By adding an element, which we denote by $∞$, and introducing the convention that $x < ∞$ for all $x ∈ X$, we will assume that X has a maximum with respect to $\le$.

We define $ω_1 = \text{min} \{ x ∈ X : \{y ∈ X : y < x \} \ \text{is uncountable}\} $

Clearly such a $ω_1$ exists. we define $0 = \text{min} \ X$ and intervels $[0,x], [x,y), (a,b]$ etc. in the usual sense.

Here are my questions:

(1)For any countable $A ⊂ [0, ω_1)$ there is an $x < ω_1$ so that $A ⊂ [0, x]$.(I don't know how to make use of the countability and uncountablity here)

(2)Equipped with the topology generated by open intervals, $[0, ω_1]$ is compact.

(3)A famous application of the first uncountable ordinal is to find an example of a Borel measure (with respect to the topology in (2)) that is not Radon. So how to construct a finite Borel measure $µ$ on $[0, ω_1]$ which is not a Radon measure?

Any solutions or elementary references will be appreciated!

No One
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  • The first two questions have been asked at least once or twice on the site before. In general you should limit yourself to one question per thread (unless they are very closely related, which is not quite the case here). – Asaf Karagila Apr 14 '15 at 20:15
  • (2) is true of any closed interval of ordinals $[0, \alpha]$ (see answer by Brian M. Scott that overlapped with me typing this comment). – Rob Arthan Apr 14 '15 at 20:31

1 Answers1

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For the first problem, observe that $[0,a]$ is countable for each $a\in A$. Let $B=\bigcup_{a\in A}[0,a]$.

  • Show that $B$ is countable.
  • Show that for each $b\in B$, $[0,b]\subseteq B$.
  • Conclude that $B$ is a countable initial segment of $[0,\omega_1)$ and therefore cannot be all of $[0,\omega_1)$.

For the second problem, observe that it suffices to show that every open cover of $[0,\omega_1]$ by open intervals has a finite subcover. Let $\mathscr{U}$ be such an open cover. There is some $U_0\in\mathscr{U}$ such that $\omega_1\in U_0$, so $U_0=(a_0,\omega_1]$ for some $a_0<\omega_1$. Now argue that there must be some $U_1\in\mathscr{U}$ containing $a_0$, so $U_1=(a_1,b_1)$ for some $a_1<a_0$ and $b_1>a_0$. Continuing in this manner, you get a strictly decreasing sequence $\langle a_0,a_1,\ldots\rangle$. Now use the fact that the ordering of $[0,\omega_1]$ that we’re considering is a well-order.

For the third question, consider the $\{0,1\}$-valued measure that assigns measure $1$ to a Borel set $B$ if and only if $B$ contains an uncountable, closed subset of $[0,\omega_1]$.

Asaf Karagila
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Brian M. Scott
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