Examples about compactness

Question

Compactness implies countably compactness which in turn implies limit-point compactness.

Sequentially compactness implies limit point compactness.

$Z_{+} \times \{0,1\}$ with two-point indiscrete space $\{0,1\}$ is limit-point compact but not sequentially compact.

First uncountable ordinal is sequentially compact, countably compact, but not compact.

Can anybody give me any examples for

1. Sequentially compact but not countably compact space (hence limit-point compact but not compact)

2. Countably compact but neither compact nor sequentially compact space.

?

Thanks in advance.

Answer 1

Actually, sequentially compact implies countably compact, so no example for (1) exist. The proof is actually quite straightforward:

Assume that $X$ is not countably compact. Then there exists a countable open cover $X = \bigcup_{n=1}^\infty U_n$ that has no finite subcover.

For each $m$, let $x_m \in X \setminus \bigcup_{n=1}^m U_n$. The sequence $x_m$ cannot have a limit point $x \in X$, otherwise it would be true that $x \in X \setminus (\bigcup_{n=1}^m U_n)$ for every fixed $m$, which implies that $x \in X \setminus (\bigcup_{n=1}^\infty U_n) = \emptyset$. Therefore, the sequence $x_m$ has no convergent subsequence.

As for (2), there's an example in the book Counterexamples in Topology, by Steen and Seebach (example 106): Let $\omega_1$ be the smallest uncountable ordinal with the order topology, and let $I = [0,1]$. Then $X = \omega_1 \times I^I$ is countably compact (because $\omega_1$ is sequentially compact and $I^I$ is compact) but neither sequentially compact nor compact (these properties are preserved by the projections).

Answer 2

Sequential compactness implies countable compactness, so you can’t get a countably compact space that’s not sequentially compact. You can, however, get a limit point compact space that’s not compact: for each $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let $\tau=\{\Bbb N\}\cup\{U_n:n\in\Bbb N\}$. Then $\tau$ is a $T_0$ topology on $\Bbb N$, and $\{U_n:n\in\Bbb N\}$ is an open cover of $\Bbb N$ with no finite subcover, but if $\varnothing\ne A\subseteq\Bbb N$, then every $n\in\Bbb N$ satisfying $n>\min A$ is a limit point of $A$.

For your second example, let $D$ be the discrete two-point space, and let $X=D^{\wp(\Bbb N)}\times\omega_1$. Since $\omega_1$ is not compact, neither is $X$, and since $D^{\wp(\Bbb N)}$ is not sequentially compact, neither is $X$. However, $D^{\wp(\Bbb N)}$ is compact, and $\omega_1$ is countably compact, so $X$ is countably compact. This space has cardinality $2^{2^\omega}$; there are more complicated examples of cardinality $2^\omega$, and in some models of set theory there are smaller examples yet.