32
$\begingroup$

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

$\endgroup$
4

2 Answers 2

Reset to default
17
$\begingroup$

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
$\endgroup$
3
  • $\begingroup$ @ 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. $\endgroup$
    – math
    Jun 3, 2012 at 8:20
  • 1
    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Sep 16, 2012 at 0:50
  • $\begingroup$ $I^I$ 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). $\endgroup$
    – Robert Furber
    Jun 13, 2020 at 10:58
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Didn't you post this answer somewhere else today? $\endgroup$
    – Asaf Karagila ♦
    May 11, 2013 at 23:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .