An example for uncountable compact space

Question

Would someone please give an example of a space which is compact but not countably compact space?

Is my example right? : suppose there exist a collection of sets ${\{S_i}\}$ for all $i\in \mathbb Q^*$ ($\mathbb Q^*$ are the set irrational numbers) which covers a set $A$; if $A$ is also covered by limited number of sets from ${\{S_i}\}$ (bijective to ${\{R_j}\}_{j=1}^N$ and $N\ne \infty$ then $A$ is compact but not countably compact space.

Answer 1

The space $\omega_1$ with the order topology is probably the simplest example. The simplest example that I know that does not require any knowledge of transfinite ordinals is the following one, which requires only a very basic knowledge of countability.

Let $A$ be any uncountable set, and for each $\alpha\in A$ let $D_\alpha=\{0,1\}$ with the discrete topology. Let $X=\prod_{\alpha\in A}D_\alpha$ with the product topology. Each point $x\in X$ is a function from $A$ to $\{0,1\}$; for convenience I’ll write $x_\alpha$ instead of $x(\alpha)$, so that $x=\langle x_\alpha:\alpha\in A\rangle$. Note that $X$, being a product of compact Hausdorff spaces, is a compact Hausdorff space.

For each $x\in X$ let $\operatorname{supp}(x)=\{\alpha\in A:x_\alpha=1\}$; I’ll call this set the support of $x$. Finally, let $Y=\{x\in X:\operatorname{supp}(x)\text{ is countable}\}$. (Note that countable means finite or countably infinite.)

Let $p\in X\setminus Y$ be such that $p_\alpha=1$ for each $\alpha\in A$.

• Show that $p\in\operatorname{cl}_XY$ and hence that $Y$ is not closed in $X$. Conclude that $Y$ is not compact. (With no more work you can actually show that $Y$ is dense in $X$.)

It remains to show that $Y$ is countably compact. Since $Y$ is Hausdorff, and hence $T_1$, it suffices to show that $Y$ is limit point compact, i.e., that every infinite subset of $Y$ has a limit point in $Y$. Suppose that $S$ is an infinite subset of $Y$. Then we may choose distinct points $y^{(n)}\in S$ for $n\in\Bbb N$. Let

$$A_S=\bigcup_{n\in\Bbb N}\operatorname{supp}x^{(n)}\;;$$

$A_S$ is the union of countably many countable sets, so $A_S$ is countable, and $x_\alpha^{(n)}=0$ for each $n\in\Bbb N$ and $\alpha\in A\setminus A_S$.

Let $Z=\prod_{\alpha\in A_S}D_\alpha$. For $n\in\Bbb N$ let $z^{(n)}\in Z$ be defined by $z_\alpha^{(n)}=y_\alpha^{(n)}$ for each $\alpha\in A_S$. (Thus, $z^{(n)}$ is the projection of $y^{(n)}$ to the subproduct $Z$.) $Z$ is the product of compact spaces, so it’s compact, and the infinite set $\{z^{(n)}:n\in\Bbb N\}$ therefore has a limit point $z\in Z$. Define $y\in Y$ by

$$y_\alpha=\begin{cases} z_\alpha,&\text{if }\alpha\in A_S\\ 0,&\text{otherwise}\;. \end{cases}$$

Note that $y$ definitely is in $Y$, since $\operatorname{supp}(y)\subseteq A_S$, and $A_S$ is countable.

• Verify that $y\in\operatorname{cl}_S\{y^{(n)}:n\in\Bbb N\}\subseteq\operatorname{cl}_YS$ and conclude that $Y$ is countably compact.