# compactness / sequentially compact

I'm looking for two examples:

1. A space which is compact but not sequentially compact
2. A space which is sequentially compact but not compact

Explanations why the spaces are compact / not compact and sequentially compact / not sequentially compact would be appreciated. A reference would also be appreciated. So the conclusion would be, that there's no equivalence in general. Of course they are equivalent in a metric space.

math

## 2 Answers

The following examples are from $\pi$-Base, a searchable database of Steen and Seebach's Counterexamples in Topology.

(Click on the following links to learn more about the spaces.)

• Stone-Cech Compactification of the Integers
• Uncountable Cartesian Product of Unit Interval ($I^I$)
• An Altered Long Line
• $[0, \omega_1)$ ($\omega_1$ is the first uncountable ordinal)
• The Long Line
• Tychonoff Corkscrew
• @ Austin Mohr: My first attempt was Counterexamples in Topology. Without knowing that the Cartesian product of unit interval and $[0,\omega_1)$ are examples, it's hard to find something in there. So thanks for your answer. – math Jun 3 '12 at 8:20
• There are tables at the back of the book, where you can find such examples with patience. And isn't there a Venn diagram in the section on compactness at the beginning? Of course, the automated Spacebook is easier. – GEdgar Sep 16 '12 at 0:50
• @Austin Mohr I am interested how to prove on uncountable Cartesian product of unit interval is not sequentially compact – math Mar 17 '13 at 22:12
• @AustinMohr: This link is not work. Could you please check for me. Thank you so much! – user52523 Nov 2 '13 at 12:06
• $I^I$ is correct, but it is wrong to characterize this as "uncountable cartesian product of unit interval" - we need the product to be of cardinality at least the continuum. As explained by KP Hart, it is consistent with ZFC that $[0,1]^{\aleph_1}$ is sequentially compact (necessarily with the continuum hypothesis being false). – Robert Furber Jun 13 '20 at 10:58

Example 1 with proof: Stone-Čech Compactification of the Integers $$\beta \omega$$

Proof: It is compact obviously. We will prove that $$\beta\omega$$ is not sequentially compact. Note that every infinite set in $$\beta\omega$$ has $$2^\mathfrak c$$ cluster points, hence the only convergent sequences in $$\beta\omega$$ are those which are eventually constant; therefore if $$X$$ is a subspace of $$\beta\omega$$ and $$X$$ is sequentially compact, then $$X$$ is finite. So $$\beta\omega$$ cannot be sequentially compact.

• Didn't you post this answer somewhere else today? – Asaf Karagila May 11 '13 at 23:49