A problem about compact order topological space This problem is from terrytao.wordpress.com/books/analysis-ii,Errata to the second edition (hardcover),P.390,Exercise 13.5.8 .I have difficulty proving this,appreciate any help!
Show that there exists an uncountable well-ordered set $\omega_1+1$ that has a maximal element $\infty$, and such that the initial segments $\{ x \in \omega_1+1: x < y \}$ are countable for all $y \in \omega_1+1 \backslash \{\infty\}$.   (Hint: Well-order the real numbers, take the union of all the countable initial segments, and then adjoin a maximal element $\infty$.)  If we give $\omega_1+1$ the order topology, show that $\omega_1+1$ is compact; however, show that not every sequence has a convergent subsequence.
 A: The set $\omega_1+1$ constructed in the exercise is order-isomorphic to the ordinal of the same name.
To show that $\omega_1+1$ is compact, you need to use the well-ordering principle.  In any open cover of $\omega_1+1$, $\infty$ (usually called $\omega_1$) must be covered by some open set.  This must contain some open interval covering $\infty$, with lower endpoint $\alpha$, say.  Then move down to $\alpha$ and look at the open set covering it, and so on.  By the well-ordering principle, this process must terminate in a finite number of steps.  Collecting the open sets you came across then gives you a finite subcover of the original open cover.
The other part of the exercise is incorrect.  $\omega_1+1$ is sequentially compact (meaning that any infinite sequence in $\omega_1+1$ has a convergent subsequence.)  To see this, take an infinite sequence in $\omega_1+1$. By the well-ordering principle, this sequence has a smallest member, $x_0$, say.  The members of the sequence coming after $x_0$ then also have a smallest member, $x_1$, which must be at least as large as $x_0$.  Similarly, the members of the sequence after $x_1$ have a smallest member $x_2\ge x_1$, and so on.  This gives you an infinite nondecreasing subsequence $x_0\le x_1\le x_2\le \cdots$.  Now, let $S:=\{z\mid z\ge x_i \text{ for all } i\}$, which is a nonempty set as it contains $\infty$.  Then the sequence $(x_i)$ converges to the minimum element of $S$.
Although $\omega_1+1$ is not such a space, there are topological spaces which are compact but not sequentially compact.  An example of such a space is $2^{2^\omega}$, where as usual $2=\{0,1\}$ and $\omega=\{0,1,2,3,\dots\}$, and the space is given the product topology of uncountably many copies of the discrete topology on $\{0,1\}$. 
This space is compact by Tychonoff's theorem but it's easy to prove that the evaluation sequence 
$$
e_0, e_1, e_2, \dots, \qquad \qquad e_i(q)=q(i) \ \ \text{for all } q\in 2^{\omega} \text{ and } i\in\omega, 
$$
has no convergent subsequence.
