# 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.

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.

• In tao's exercise,I think $\omega_1\neq \infty$,and I think tao didn't take $\omega_1+1$ as an ordinal.See his hint. – Luqing Ye Mar 15 '13 at 4:06
• This construction gives a linear ordering which is order-isomorphic to the usual ordinal $\omega_1+1$. When you take the union of the countable initial segments of a well-ordering of $\Bbb R$, this gives a well-ordered set which is order-isomorphic to $\omega_1$. Adding a maximal element $\infty$ then does the same thing as taking the successor of $\omega_1$, giving $\omega_1+1$. – David Moews Mar 15 '13 at 4:07
• Thanks for your explaination.But then I can't understand this "Then move down to $\alpha$ and look at the open set covering it, and so on."."And so on" makes me confusing.How can that prove it is compact? – Luqing Ye Mar 15 '13 at 4:25
• By the well-ordering principle, the process must terminate in a finite number of steps. That gives you a finite subcover. – David Moews Mar 15 '13 at 4:27
• If it went on forever, you would have an infinite decreasing sequence $\alpha_1>\alpha_2>\alpha_3>\cdots$. This is impossible, since then the set $\{\alpha_i\}$ would have no least element, which contradicts the well-ordering principle. – David Moews Mar 15 '13 at 4:32