Sub-base theorem Is this proof correct? I have the following proof of this theorem, I think the proof is correct but a doubt arose when using Zorn's lemma.
Theorem: Let S be a subbase of a topological space X. Prove that if every cover of X consists of elements of S has a finite sub-cover then X is compact.
Proof:
Suppose that X is not compact. Let U be an open cover of X consisting of basic sets that do not have a finite sub-cover.
Let us consider the family
$F = \{U: U$ is an open cover of X that does not contain finite sub-covers $\}$.
This set is partially ordered and each sub-collection of this that is fully ordered has an upper bound.(Why? How is this proven?)
By Zorn's lemma there exists $V \in F$ a maximal sub-cover that has no finite sub-cover.
Let $U_{\alpha}$ be a basic element such that $U_{\alpha} \in V$, then there exists $S_{\alpha_{1}},...,S_{\alpha_{r}} \in S$ (subbasic) such that $U_{\alpha}= S_{\alpha_{1}} \cap...\cap S_{\alpha_{r}}$.
Suppose that $S_{\alpha_{i}} \notin V$ for all $i = 1, ... r$, then $V \cup \{S_{\alpha_{i}}\} \notin F$ since by hypothesis we have that every cover formed by elements of S has a finite sub-cover that covers X. This tells us that $X-S_{\alpha_{i}}$ is covered by finite elements of V, but
$X-U_{\alpha}=(X-S_{\alpha_{1}}) \cup...\cup(X-S_{\alpha_{r}})$ so each $X-U_{\alpha}$ is covered with a finite number of elements of V. And note that $X=(X-U_{\alpha}) \cup U_{\alpha}$ then $X$ will be covered with a finite number of elements of V, which is a contradiction.
If each $U_{\beta} \in  V$ then there exists a subbasic $S_{\beta}$ such that $U_{\beta} \subset S_{\beta}$ then X also has a cover of subbasics for X and by hypothesis we have a finite sub-cover of subbasics that cover X, this implies that there are finite bases $U_{\alpha}$ to cover X, which contradicts the choice of V. Therefore X is compact.
Is this proof correct?
I thank you in advance for the help.
 A: First the issue of the application of Zorn's lemma (or some other for of AC):
Zorn will work fine, but a better fit here is the equivalent form of AC known as the Teichmüller–Tukey lemma:
The collection $\mathscr{U}=\{\mathcal{U}: \mathcal{U} \text{ is an open cover of } X \text{ without a finite subcover}\}$ is a collection of finite character, as can trivially be checked (see the definition on the linked Wikipedia page): a cover is a member of $\mathscr{U}$ iff every finite subset is a member of $\mathscr{U}$ as well.
By assumption of $X$ being non-compact, $\mathscr{U}$ is non-empty and so the lemma applies: there is some maximal (for inclusion) member of $\mathscr{U}$, call it $\mathcal{O}$.
Now make the following observations (always true for such a maximal cover; no further assumption is needed):

*

*If $O \notin \mathcal{O}$, then $O^\complement$ has a finite cover by elements of $\mathcal O$ (follows from the fact that by maximality $\mathcal{O} \cup \{O\}$ has a finite subcover).

*If $O \in \mathcal{O}$ then any open subset of $O$ is also in $\mathcal{O}$ (by a similar maximality argument).

*$\mathcal{O}$ is closed under finite unions.

*If $O_1, \ldots, O_n$ are open and $O:=\bigcap_{i=1}^n O_i \in \mathcal{O}$, then for some $j\in \{1,\ldots,n\}$ we have $O_j \in \mathcal{O}$: (this is an argument you also give, really) Proof: otherwise $O_i^\complement$ would have a finite subcover from $\mathcal{O}$ for all $i$ and their union (i.e. of these subcovers) would be a finite subcover of $O^\complement$ which would give $\mathcal{O}$ a finite subcover (after adding $O$), contradiction.

So we've essentially shown that $\mathcal{O}$ is an ultra-ideal in the open sets... (you can forget about that if you've never seen ideals or filters, just an aside).
Now we can finally use the subbase $\mathcal{S}$ from the assumption (note that we don't use a subbase until the very end!):

Claim: $\mathcal{O} \cap \mathcal{S}$ is a cover of $X$.

Let $x \in X$. As $\mathcal{O}$ is an open cover we have some $O \in \mathcal{O}$ such that $x \in O$. As the finite intersections of subbase elements is a base for $X$, we have $S_1,\ldots, S_n \in \mathcal{S}$ such that $$x \in \bigcap_{i=1}^n S_i \subseteq O\tag{1}$$.
The second property followed by the fourth one implies that $\bigcap_{i=1}^n S_i \in \mathcal{O}$ so some $S_j, j\in \{1,\ldots,n\}$ is in $\mathcal{O}$. But then $x \in S_j \in \mathcal{S} \cap \mathcal{O}$ and $x$ is covered by all subbasic elements of $\mathcal{S}$.
The assumption on $\mathcal{S}$ then tells us that we have a finite subcover of $\mathcal{S} \cap \mathcal{O}$ and this is then a finite subcover of $\mathcal{O}$, the final contradiction.
So structure of the proof: The assumption of non-compactness of $X$ tells us $\mathscr{U}$ is non-empty and so AC (or Teichmüller-Tukey) gives us a maximal element which leads to a contradiction with the assumption on $\mathcal{S}$.

Appendix: How to use Zorn instead of the T-T lemma.
I can imagine that many texts do not cover the Teichmüller-Tukey lemma at all, but teach the quite general Zorn lemma instead. In the case at hand this will yield a proof as well:
The poset is $P:=\mathscr{U}$ from above, so all open covers without a finite subcover. Assuming that $X$ is not compact tells us in essence that this set is non-empty (which is needed to apply the lemma, obviously: no elements then no maximal element). The order on this set $\le_P$ is just inclusion of covers: $\mathcal{U} \le_P \mathcal{V} \iff \mathcal{U} \subseteq \mathcal V$. This clearly obeys the poset axioms (inclusion on a power set is, and this is a restriction to special families..) Now the assumption on $(P, \le_P)$ to apply Zorn is that the partial order is "inductive": every chain (subset of $P$ that happens to be linearly ordered by $\le_P$) $P' \subseteq P$ has an upper bound in $P$. So let's show that.
Let $P':=\mathscr{V} \subseteq \mathscr{U}=P$ be a non-empty chain in $P$.
So this is a set of open covers (all of which have no finite subcover) so that for any two covers $\mathcal{U}_1. \mathcal{U}_2 \in P'$ we either have $\mathcal{U}_1 \subseteq \mathcal{U}_2$ or $\mathcal{U}_2 \subseteq \mathcal{U}_1$. Now define $\mathcal{U}_{P'}:= \bigcup \{\mathcal{U}: \mathcal{U} \in \mathscr{V}\}$; being a non-empty union of open covers it is still an open cover of $X$, obviously and clearly (a union is a superset of each of the sets in a union) we have $\forall \mathcal{U}\in \mathscr{V}: \mathcal{U} \le_P \mathcal{U}_{P'}$, so this would be an upper bound for  $P'$ if only $\mathcal{U}_{P'} \in P$ so iff it has no finite subcover. And this is quite easy to see: suppose $\mathcal{U}_{P'}$ had a finite subcover $\{U_1,\ldots, U_m\}$ for some $m \in \Bbb N$: for each $i \in \{1,\ldots,m\}$: as $U_i \in \mathcal{U}_{P'}$ we have some $\mathcal{U}_i \in \mathscr{V}$ so that $U_i \in \mathcal{U}_i$. So this way we have finitely many covers $\mathcal{U}_1, \ldots ,\mathcal{U}_m\}$ in a linearly ordered (by inclusion) set $\mathscr{V}$. It follows that there is a maximum among these (a finite subset of a linearly ordered set has a maximum and a minimum; easy first year lemma for an induction proof e.g.) say $\mathcal{U_i}\subseteq \mathcal{U}_l$ for all $i \in \{1,\ldots,m\}$ and some $l \in \{1,\ldots,m\}$. It follows easily that in that case
$\{U_1,\ldots, U_m\} \subseteq \mathcal{U}_l$, but this contradicts that $\mathcal U_l \in \mathscr{U}$: it cannot have a finite subcover... This contradiction shows that indeed $\mathcal{U}_{P'} \in \mathscr{V}$ and thus is the required upperbound for $P'$ and we're done showing that $(P, \le_P)$ has a maximal element for inclusion, giving the same $\mathcal{O}$ as T-T gave us earlier. You can here also see the contours of a general proof that Zorn implies the T-T lemma...
