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?

  • $\begingroup$ @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. $\endgroup$ – Reveillark Mar 4 at 3:55
  • $\begingroup$ Q* is ill defined because $\Omega$ is not a set. $\endgroup$ – William Elliot Mar 4 at 3:55
  • $\begingroup$ @WilliamElliot This is the definition I am given. $\endgroup$ – user312437 Mar 4 at 4:24

Another answer to address the proof the OP proposed.

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 $\Omega$).

  • $\begingroup$ Thank you very much. This both answers my question and clears up whatever confusion I had regarding finding a finite cover. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – user312437 Mar 4 at 16:13
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – user312437 Mar 4 at 16:15
  • $\begingroup$ @DerekAdams see my second answer for that argument. $\endgroup$ – Henno Brandsma Mar 4 at 20:23

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