Well-ordered set with greatest element is compact

Question

Let $X$ be a well-ordered set with a greatest element $\alpha$. We consider all sets of the form $]x,y]$ where $y \in X$ and $x$ is either another element of $X$ or the symbol $\leftarrow$ (the interval $]\leftarrow,y[$ is the set of all elements $<y$). They generate a topology $T$. I would like to prove that $(X,T)$ is a compact space.

It is obviously separated. Also, any closed interval $[x,y]$ is a closed set and if $[x_i,y_i]$ ($i \in I$) is some family of closed intervals such that the intersection of a subfamily is non-empty, then it is easy to see that the intersection of all $[x_i,y_i]$ is non empty (look at $\sup x_i$ and $\min y_i$). Note that since $X$ has a greatest element, any non-empty set has a supremum indeed.

We must prove the same fact when closed intervals are replaced with general closed sets. This I can't do. Could you give me some advice?

Thanks.

PS. I'm using the french notation for intervals, i.e. $[x,y]$ is the set of all $z$ such that $x \leq z \leq y$, etc.

Answer 1

If the conclusion fails, then there is, in $X$, a first element $\beta$ such that $]\leftarrow,\beta]$ is not compact. Let $C$ be an open cover of $]\leftarrow,\beta]$ with no finite subcover, and let $U$ be an element of $C$ that contains $\beta$. Of course, $U$ doesn't include all of $]\leftarrow,\beta]$, so, by definition of the topology, it includes $]\gamma,\beta]$ for some $\gamma<\beta$. By the minimality of $\beta$, finitely many elements of $C$ cover $]\leftarrow,\gamma]$. These finitely many together with $U$ cover $]\leftarrow,\beta]$; contradiction.

Answer 2

I'd use the fact that $F\subset X$ is closed if and only if for each strictly increasing (possibly transfinite) sequence $(x_\alpha)_{\alpha<\lambda}$ with entries in $F$ we have

$$\big( \lim_{\alpha<\lambda} x_\alpha = \big) \sup_{\alpha<\lambda}x_\alpha\in F.$$

Now, if $F$ is closed in $X$, by the above criterion $F$ has the greatest element. It is well-ordered itself so there is an increasing bijection $\Theta$ from $F$ onto some ordinal which itself has the greatest element. But for such interval you've already proved the theorem and $\Theta$ is also a homeomorphism.

Answer 3

I would try this: Given a family of closed subsets where any finite intersection is nonempty, add all such finite intersections to the family, so you have a family of nonempty, closed subsets which is closed under finite intersections. Any $F$ in the family has a smallest and a largest element. Consider the supremum of all the $\min F$, or the infimum of all the $\max F$. Either should be in the intersection of all $F$.

Answer 4

Prove by induction on $\beta$ that $L(\beta) = \{x: x \le \beta \}$ is compact, for all ordinals $\beta$.

This is clear for $\beta = 0$, where $L(0) = \{0\}$ and if $\beta+1$ is a successor, then $L(\beta+1) = L(\beta) \cup \{\beta+1\}$, so if $L(\beta)$ is compact, so is $L(\beta+1)$.

So assume $L(\alpha)$ is compact for all $\alpha < \beta$ and $\beta$ is a limit ordinal. Let $(U_i), i \in I$ be an open cover of $L(\beta)$. Some $U_j$ covers the point $\beta \in L(\beta)$, and so for some $\alpha < \beta$ we have that $(\alpha, \beta] \subset U_j$. Then as $L(\alpha)$ is compact, it is covered by finitely many $U_i$, and adding $U_j$ gives us the required finite subcover. So $L(\beta)$ is compact.

Answer 5

This is quite similar to Andreas Blass' answer, but as a positive proof instead of a negative proof. I will use standard ordinals notation, so $0$ is the least element, and a successor ordinal means a point that has a predecessor in the order. I will also use $(x,y)$ to denote the open interval from $x$ to $y$.

Suppose that $\mathcal U$ is an open cover of $[0,\eta]$.

Let $\alpha_0=\eta$. There is at least one $U_0\in\mathcal U$ such that $\alpha_0\in U_0$. Since $U_0$ can be written as the union of open intervals, either $U_0=[0,\eta]$ (which is open, as it's the entire space), or there is a least $\alpha_1\notin U_0$ such that $(\alpha_1,\eta]\subseteq U_0$. Let $U_1$ be some open set in $\mathcal U$ such that $\alpha_1\in U_1$. By induction define $\alpha_{n+1}$ the least ordinal such that $\alpha_{n+1}\notin\bigcup_{i\leq n}U_i$ but $(\alpha_{n+1},\eta]\subseteq\bigcup_{i\leq n}U_n$. If there is no such ordinal we halt.

Now we have that $\ldots<\alpha_n<\ldots<\alpha_1<\alpha_0$. This is a decreasing sequence of ordinals, so it has to be finite. Therefore there must be some $k$ such that $\bigcup_{i\leq k}U_i=[0,\eta]$ as wanted.