# Can't the first (/second) transinfinite ordinal replace the first (/second) uncountable ordinal in several counterexamples?

An example of a sequentially compact but not compact space is $\omega_1$.

Indeed, any sequence of countable ordinals either has infinite elements below some countable ordinal, or has an ascending subsequence. In the latter case, the limit of the ascending subsequence is the union of its elements, which is a countable union of countable sets $\Rightarrow$ a countable set $\Rightarrow$ an element of $\omega_1$.

Also, $\omega_1$ is not compact, as we can cover it by opens $[0,a)$, with $a$ a countable ordinal.

But wouldn't the same argument also go for $\omega_0$, the first transfinite ordinal?

And isn't $\omega_0+1$ compact, in the same way as $\omega_1+1$ is?

Another counterexample that makes use of the first uncountable ordinal is of a measure that is neither inner nor outer regular. Let $\mu$ be the (Borel) measure on $\omega_1+1$ that assigns $1$ to unbounded intervals and $0$ to bounded intervals. Again, couldn't we replace $\omega_1+1$ by $\omega_0$?

No - $\omega$ is definitely not sequentially compact. (Consider the sequence 1, 2, 3, 4, . . .) This is one place where uncountable ordinals (actually, ordinals of uncountable cofinality) are relevant.
As to the measure example, if we use $\omega$ then a measure 1 set is a union of countably many measure 0 sets.
• Thanks! After I did some research on what you said, I would like to point out, however, that ordinals of uncountable cofinality are not a general case of uncountable ordinals, as I interpreted you on a first reading. $\omega_\omega$ is an uncountable ordinal of countable cofinality. I did a small edit on your post to clarify that. I hope it didn't hurt. – Rodrigo Jan 21 '16 at 14:59