# An example for uncountable compact space

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.

• You seem to have the implication backwards: a compact space is countably compact, while a countably compact space needn't be compact. Wikipedia gives an example: the first uncountable ordinal $\omega_1$ with its order topology is countably compact but not compact. Since metrizable spaces are compact if and only if they are countably compact, this is about as simple as any example could ever be. – Ian Jul 22 '15 at 10:25
• @DavidMitra - I know the definitions. I am trying to figure out their differences by example.. – L.G. Jul 22 '15 at 10:28
• It should be evident that compact implies countably compact. For the converse, see this. – David Mitra Jul 22 '15 at 10:29
• @Ian - "a compact space is countably compact, while a countably compact space needn't be compact", would you please give examples to show why is so? Wiki's examples are just examples by them I can't figure out the statement you wrote. Thank you – L.G. Jul 22 '15 at 10:33
• @DavidMitra - thank you for the link; I don't about $\omega_1$ topology or carnality or order topology. Is there some elementary (undergraduate level) example for why countably compact space may not be compact? Is my example correct? Thank you – L.G. Jul 22 '15 at 10:41

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.