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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.


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4 – user38268 Jun 1 '12 at 12:34
See also: – t.b. Jun 1 '12 at 12:47
...and the canonical reference for answering such questions quickly and reliably is Steen and Seebach, Counterexamples in Topology. – t.b. Jun 1 '12 at 12:57
up vote 7 down vote accepted

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.)

For compact but not sequentially compact:

  • Stone-Cech Compactification of the Integers
  • Uncountable Cartesian Product of Unit Interval ($I^I$)

For sequentially compact but not compact:

  • An Altered Long Line
  • $[0, \omega_1)$ ($\omega_1$ is the first uncountable ordinal)
  • The Long Line
  • Tychonoff Corkscrew
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@ 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

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.

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

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