# Showing the uncountable well-ordered set with a largest element is compact.

I'm having some trouble proving what is said above. My work is as follows:

Let $$Q$$ be the uncountable well-ordered set and let $$\Omega$$ be such that $$\Omega > q$$ for all $$q \in Q$$. I am trying to show that $$Q^* = Q \cup \Omega$$ is compact. To start, assume that $$Q^*$$ is not compact and that $$O$$ is an open cover of $$Q^*$$ that has no finite subcover. For each $$q \in Q^*$$, let $$T_q = \{p \in Q^* \ | \ p \le q\}$$. Let $$C = \{p \in Q^* \ | \ T_p \text{ cannot be covered by finitely many sets from }O\}$$. $$C$$ is nonempty by assumption, and so let $$m$$ be the minimal element of $$C$$ (which cannot be the minimal element of $$Q$$). $$T_m$$ is the smallest section that cannot be covered by finitely many intervals, but for all $$q < m$$ $$T_q$$ can be covered by finitely many subcovers of $$O$$. I am struggling here at this final step. If $$O_q$$ is a finite subcover of $$O$$ that covers $$T_q$$, how can I construct a finite subcover of $$T_m$$ and show a contradiction? Initially, I went with $$O_q \cup [q,c')$$, for $$c'$$ the successor of $$c$$, but there's no guarantee that this open set is a set of $$O$$. Any hints would be appreciated.

Edit: I am not asking for a proof of the aforementioned statement, but commentary/criticism on my partial proof above (otherwise I would not have included it). That is, how do we construct a finite subcover of $$T_m$$ given the material above? If we cannot, why?

• @WilliamElliot Whoopsie daisy :P You're totally right, of course. Being well ordered is unnecessarily strong though, complete and linearly ordered is sufficient, which showcases what goes wrong with the rationals. – Reveillark Mar 4 at 3:55
• Q* is ill defined because $\Omega$ is not a set. – William Elliot Mar 4 at 3:55
• @WilliamElliot This is the definition I am given. – user312437 Mar 4 at 4:24

To set the stage: So we have the set $$X=\Omega \cup \{\Omega\}$$ where $$\Omega$$ is the smallest uncountable ordinal. $$\Omega \in X$$ is the maximum of $$X$$. $$X$$ has the order topology, which has as a base all sets of the following forms

• $$[0,x), x \in X$$ (the only basic open sets that contain $$0$$),
• $$(x,y): x < y \in X$$,
• $$(x,\Omega], x \in X$$ (the only basic open sets that contain $$\Omega$$).

Now let $$\mathcal{U}$$ be an open cover of $$X$$. Suppose that it does not have a finite subcover (striving for a contradiction). Define $$M = \{x: [0,x] \text{ does not have a finite subcover by } \mathcal{U}\}$$

and note that, as $$\Omega \in M$$, $$M \neq \emptyset$$, so $$m=\min(M) \in X$$ exists. Also $$0 \notin M$$, clearly.

First case: $$m=\Omega$$. Then as we have a cover of $$X$$ some $$U_0 \in \mathcal{U}$$ exists with $$\Omega \in U_0$$, and so there is some $$x_0 \in X$$ such that $$(x_0,\Omega] \subseteq U_0$$. As $$x_0 < m=\min(M)$$, $$[0,x_0]$$ has a finite subcover by elements from $$\mathcal{U}$$ (else it would contradict the minimality of $$m$$) and adding $$U_0$$ to these, shows that we also have a finite subcover for $$[0,m]$$, a contradiction, as then $$m \notin M$$ at all.

Second case (really the same) $$m < \Omega$$ (or equivalently $$0< m \in \Omega$$): then again we find $$U_0$$ from $$\mathcal{U}$$ such that $$m \in U_0$$ and a basic $$(x_0, x_1)$$ with $$m \in (x_0, x_1) \subseteq U_0$$. Again we have a finite subcover for $$[0,x_0]$$ and we add $$U_0$$ to cover $$[0,m]$$ again by a finite subcover and so $$m \notin M$$ contradiction.

We could have combined the above two cases by a priori showing that in a well-order the order topology has as a base all sets of the form $$(x,y], x < y \in X$$ and use only sets of that form in our argument.

But the minimality argument is a valid one. The max is needed for non-emptyness of $$M$$, that's why the argument fails for the non-compact space $$\Omega$$ itself. This clearly has a cover $$\{[0,x): x \in \Omega\}$$ without even a countable subcover (as all countable subsets of $$\Omega$$ have an upperbound in $$\Omega$$).

• Thank you very much. This both answers my question and clears up whatever confusion I had regarding finding a finite cover. – user312437 Mar 5 at 17:50

A well ordered set with a maxinum element is a complete linear order.
Showing your set is compact is a result of showing complete linear orders with the order topology are compact.
To prove that, the Alexandroff subbase theorem is used, namely:
if every cover of a space S by subbase sets has a finite subcover, then S is compact.

Let S be a complete linear order.
Since S is complete it has a top t, and bottom b, element.
B = { [b,x), (x,t] : x in S }, the collection of all sets of the form [b,x) and (x,t] is a subbase for the order topology of S.

Now if C subset B is a cover of S, let
u = inf{ x : (x,t] in C }.
As C covers S, exists y with u in [b,y) and [b,y) in B.
Thus { [b,y), (u,t] } is a finite subcover of C.

Consequently, by the subbase theorem, S is compact.

• Have you a proof that does not rely on such strong tools? Was my idea of a proof not correct? Could it not be extended to prove the result? – user312437 Mar 4 at 16:13
• Define Q* correctly., Q $\cup$ { $\Omega$ }. But that is still wrong, a counterexample possible. $\Omega$ in Q is needed. Would you like a proof of Alexandrov's subbase theorem? – William Elliot Mar 4 at 23:25

An ordered space $$X$$ in the order topology is compact iff $$\sup(A)$$ exists for all $$A \subseteq X$$.

Now let $$X$$ be non-empty well-ordered with $$M=\max(X)$$.

If $$A\subseteq X$$, the set $$U(A)$$ of uperbounds of $$A$$ is non-empty (e.g. $$M$$ is in it) and then $$\min(U(B))= \sup(A)$$ exists by well-orderedness. So $$X$$ is compact in the order topology.

• Although I do believe you, I was curious as to whether or not my idea of a proof was correct/incorrect, and why. I am not familiar with the theorem you cite above, so I am not confident in using it in my proof. This may seem nit-picky but it is also a matter of preference. – user312437 Mar 4 at 16:15
• @DerekAdams see my second answer for that argument. – Henno Brandsma Mar 4 at 20:23