Examples about compactness 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 


*

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

*Countably compact but neither compact nor sequentially compact space.
?
Thanks in advance.
 A: 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).
A: 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.
