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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 $\omega_0$?

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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.

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  • $\begingroup$ 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. $\endgroup$ – Rodrigo Jan 21 '16 at 14:59
  • $\begingroup$ No, that's why I used the word "specifically" instead of "more generally" - ordinals of uncountable cofinality are a specific kind of uncountable ordinal. I'm okay with it this way, but the other way is correct. $\endgroup$ – Noah Schweber Jan 21 '16 at 15:18
  • $\begingroup$ Yeah, after I understood what you meant, I realized that you were right in saying it that way too. But the first time I read it, I was misled. Probably it's just my feeling for colloquial language, and I thought it could help putting it this way. $\endgroup$ – Rodrigo Jan 22 '16 at 11:52

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