# Showing Compactness

#### joypav

##### Active member
I am working on a little series of problems that my professor gave to us. We are to use the claims we've proven to show X is compact.

Definition of Perfectly Compact:
The point set M is said to be perfectly compact if and only if it is true that if G is a monotonic collection of non-empty subsets of M then there is a point p that is a point or a limit point of every every element of G.

Problem: Prove that if X is perfectly compact, then it is compact. (Hint: This will be an application of the well-ordering theorem)

Claim 0: Suppose that $\mathcal{B}$ is a basis for the topology of the Hausdorff space $X$. The space $X$ is compact iff for each covering $G$ of $X$ with $G \subset \mathcal{B}$, some finite subcollection of $G$ covers $X$.
(a simple proof, I won't type it up here)

Suppose in the following that $X$ is a perfectly compact space and $\mathcal{B}$ is a basis for the topology of $X$. And $G$ is a collection of elements of $\mathcal{B}$ covering $X$.

Claim 1: If there is a countable basis $\mathcal{B}$ for $X$ then $X$ is compact.
(I proved this in a previous post on here)

Claim 2: If there is a well-ordering of the elements of $G$ then there is an element $g$ of $G$ so that $g$ plus the element of $G$ that precede $g$ covers $X$.
[Corollary: There is a first such g.]
Proof:
Let $G$ be an open cover of $X$ that is well-ordered, with respect to, say "$\prec$".
Then,
$\forall g_\alpha , g_\beta \in G , g_\alpha \prec g_\beta$ or $g_\beta \prec g_\alpha$ and $g_\alpha = g_\beta$ iff $\alpha = \beta$. ($\alpha , \beta$ in some indexing set $I$)
Let,
$V_\alpha = \cup\left\{g_i \in G : g_i \preceq g_\alpha\right\}$
Then, if $g_\alpha \prec g_\beta \implies V_\alpha \subset g_\beta$
Let,
$U_\alpha = X - V_\alpha$

Case I: $\exists \alpha$ s.t. $X = V_\alpha$
If this is the case then we are done. Because $X = V_\alpha = \cup\left\{g_i \in G : g_i \preceq g_\alpha\right\}$ is a cover of $X$ such that $g_\alpha$ and the elements of $G$ preceding it covers $X$.

Case II: there is no such $\alpha \in I$ as in Case I
Then,
$\forall \alpha , U_\alpha \neq \emptyset$ and $\cap_{\alpha \in I}U_\alpha \neq \emptyset$
$\implies \exists x \in X, x \in \cap_{\alpha \in I}U_\alpha \implies x \in \cap_{\alpha \in I}\left(X - V_\alpha\right) \implies x \notin \cup_{\alpha \in I} V_\alpha$,
a contradiction with our definition of the $V_\alpha$'s.
$\implies$ Case I is the only possible case

Proof of Corollary:
I wasn't sure about this. Can we just define a set of all the $g$'s such that it and the elements before it cover $X$? We have just shown that such a set is nonempty, so then it would have a least element.

Claim 3: If $G$ is well-ordered with the ordering $<$, then there is no infinite decreasing sequence of elements of $G$.
Proof:
BWOC, assume such a sequence does exist, $\left\{g_i\right\}_{i=1}^\infty , g_i \in G$.
Define a set,
$S = \left\{g \in G : g \geq g_1\right\} = \left\{g_1, g_2, g_3,...\right\}$
Then $G$ well-ordered and $S \subset G \implies S$ has some minimal element, say $g_m$.
But this is a contraction with our assumption that the sequence is infinite, because $g_m$ would be the last term in the sequence.

Now, I need to show that if $X$ is perfectly compact then it is compact. But how? I am not allowed to assume that it is well-ordered, so I'm confused about how to use the claims I've proven.
Also, are my proofs for Claim 2 and Claim 3 okay?