I want to show that $[0, \omega_1]$  is compact, where $\omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals.   The tips I have seen to showing this is that $[0, \omega_1]$:   contains a maximal element there are no infinite strictly decreasing sequences of ordinals.     Take an open cover of the desired set. My thoughts are that some open set must contain $\omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
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 $&lt;$ means $\le$ but not euqual). By adding an element, which we denote by $∞$, and introducing the convention that $x &lt; ∞$ 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 &lt; 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 $x &lt; ω_1$ 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!
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