# Minimal Uncountable Well-Ordered Set

#### joypav

##### Active member
(The notation in this problem is a bit different than what I have seen elsewhere, however it is how it was stated in my class so I will keep it.)
I'd like to make sure my proof for (i) is okay, and need some clarification on the proof for (ii) that was given.

Problem:
Suppose that $X'$ is an uncountable, well-ordered set and put the order topology on $X'$. Then either

Case 1: There is a point $p$ so that $\{ x:x<p \}$ is uncountable.
Case 2: If $p \in X'$ then $\{ x:x<p \}$ is countable.

In Case 2, let $\omega_1$ denote the first point $p$ of $X'$ so that $\{ x:x<p \}$ is uncountable.
Since the two are topologically equivalent (something I'd like to show but haven't gotten to yet), consider $X = \{ x:x<\omega_1 \}$. Show that:
(i) $X$ is not compact.
(ii) $X$ is limit compact.

(i)
Claim 1: $X$ has no largest element.
BWOC, assume it does.
$\implies \exists y \in X$ so that $x \leq y, \forall x \in X$.
Consider $S = \{ x:x<y \}$.

Claim 2: $S \cup \{ y \} = X$
"$S \cup \{ y \} \subset X$"
Let $x \in S \cup \{ y \}$.
Then,
$x \in S \implies x<y \implies x \in X$
or
$x=y \in X \implies x \in X$

"$X \subset S \cup \{ y \}$"
Let $x \in X \implies x < \omega_1, y$ largest element of $X \implies x \leq y \implies x \in S \cup \{ y \}$

$\implies$ Claim 2 is true.
Because $X = S \cup \{ y \}$,
$X$ uncountable $\implies S$ is uncountable.
But this is a contradiction with the definition of $X$. (it is also worth noting that $y$ cannot be equal to $\omega_1$, by its definition)

$\implies$ Claim 1 is true.

Now, consider $G = \{ g= \left(-\infty, x\right) : x \in X \}$.
Claim: $X \subset \cup_{g \in G} g$
Let $x \in X \implies x < \omega_1$ and $\exists y \in X$ such that $x < y < \omega_1$, (because $X$ has no largest element).
Then, $x \in \left(-\infty, y \right) \subset \cup_{g \in G} g$.
$\implies$ Claim is true.
Obviously each $g$ is open, so $G$ is an open cover of $X$.
However, there does not exist a finite subcover of $G$ that covers $X$, because if there were, $X$ would have to have a largest element. (Do I need more detail?)

$\implies X$ is not compact.

(ii)
Let $A \subset X$, $A$ infinite.
Choose $B$ such that $B$ is a countably infinite subset of $A$. (How can we get such a $B$?)

Then $B \subset A \subset X, B$ countable
$\implies \exists b$ such that $b$ is an upper bound for $B$ and $b \in X$. (Is this a result of $X$ being well-ordered?)

$X$ is well-ordered $\implies \exists a_0$ that is the smallest element of $X$.
Obviously, $B \subset [a_0, b]$.

$X$ well-ordered $\implies X$ has the l.u.b. property $\implies [a_0, b]$ is compact. (I will have to prove this for myself... but it should be fine. I believe the compactness of every closed interval implies that the space has the l.u.b. property... I assume that is what is used here.)
$\implies [a_0, b]$ is limit point compact (as compactness $\implies$ limit point compactness)
$\implies B \subset [a_0, b]$ has a limit point $x \in [a_0, b] \implies x \in A$

$\implies X$ is limit point compact.