A property of compactness

Question

In http://www.sciencedirect.com/science/article/pii/S0166864107000193, Van Mill showed that a space $X$ is compact iff for any neighbourhood assignment $\{O_x: x\in X\}$ there is a compact subset $K$ of $X$ s.t. $\cup\{O_x : x \in K\} = X$. (See Theorem 2.5)

Note: A neighbourhood assignment in a space $X$ is a family $\{O_x: x\in X\}$ such that $x \in \tau(X)$ for any $x \in X$.

My question: why the following statement is correct: if $X$ is not compact then then there is an infinite cardinal $\lambda$ and a cover $\mathcal{U} = \{U_{\alpha}: \alpha < \lambda\} \subset \tau(X)$ of the space $X$ such that $\alpha < \beta$ implies $U_{\alpha} \subset U_{\beta}$ and $U_{\alpha} \neq X$ for any $\alpha < \lambda$.

It may be a silly question but I really do not understand. I only know that if $X$ is not compact then there is open cover $\mathcal{U}$ such that there is no finite subcover. Plese help me to understand this one.

Answer 1

Let $\mathcal{V}$ be any open cover of $X$ that has no finite subcover.

Then well-order this cover (like Mary-Ellen Rudin used to say "well-order everything in sight"), so using a bijection of the cover with some ordinal $\gamma$ (We can even use a cardinal number $\gamma$ of course):

$\mathcal{V} = \{U_\alpha: \alpha < \gamma \}$. Define $U_\alpha = \bigcup \{ U_\beta: \beta < \alpha\}$. If $\alpha < \beta$, then we know that $U_\alpha \subseteq U_\beta$, because we take the union of more sets.

We know by definition that $U_\gamma = X$ as we have a cover.

So define $\gamma_0 = \min \{\alpha : \alpha \text{ is an ordinal and } U_\alpha = X\}$. This is well-defined as then ordinals are well-ordered an we take the minimum of a non-empty set of ordinals. And $\gamma_0$ is infinite as $\mathcal{V}$ has no finite subcover.

Now go to a cofinal subset of $\gamma_0$ by removing duplicates (when $U_\alpha = U_{\alpha+1}$ which can happen before we reach $X$) and reindex with a cardinal, keeping the order. This satisfies the requirement.