# Example 2, Sec. 28, in Munkres' TOPOLOGY, 2nd ed.: Does every compact ordered (or well-ordered) set always have a largest element.

Here is Example 2, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:

... Consider the minimal uncountable well-ordered set $$S_\Omega$$, in the order topology. The space $$S_\Omega$$ is not compact, since it has no largest element. However, it is limit point compact. Let $$A$$ be an infinite subset of $$S_\Omega$$. Choose a subset $$B$$ of $$A$$ that is countably infinite. Being countable, the set $$B$$ has an upper bound $$b$$ in $$S_\Omega$$; then $$B$$ is a subset of the interval $$\left[ a_0, b \right]$$ of $$S_\Omega$$, where $$a_0$$ is the smallest element of $$S_\Omega$$. Since $$S_\Omega$$ has the least upper bound property, the interval $$\left[ a_0, b \right]$$ is compact. By the preceding theorem, $$B$$ has a limit point $$x$$ in $$\left[ a_0, b \right]$$. The point $$x$$ is also a limit point of $$A$$. Thus $$S_\Omega$$ is limit point compact.

Now my question is related to the following part of this example.

The space $$S_\Omega$$ is not compact, since it has no largest element.

In general, we can prove the following:

Let $$X$$ be a simply ordered set, in the order topology. If $$X$$ has no smallest element or if $$X$$ has no largest element, then $$X$$ is not compact. In other words, if a simply ordered set $$X$$ in the ordered topology is compact, then $$X$$ must have a smallest element and a largest element.

Proof:

Suppose that $$X$$ has a smallest element $$a$$ but that $$X$$ has no largest element. Then the collection $$\mathscr{A} \colon= \left\{ \ [a, x) \ \colon \ x \in X, x > a \ \right\}$$ is an open covering of $$X$$. If there were a finite sub-collection of $$\mathscr{A}$$ that also covered $$X$$, say, the collection $$\mathscr{A}_0 \colon= \left\{ \ \left[ a, x_1 \right), \ldots, \left[ a, x_n \right) \ \right\},$$ then we would have $$X = \bigcup_{i=1}^n \left[ a, x_i \right) = \left[a, x_* \right), \tag{1}$$ where $$x_* \colon= \max \left\{ \ x_1, \ldots, x_n \ \right\}.$$ But this would yield a contradiction because $$x_*$$ is an element of $$X$$, by our construction, but from (1) we would obtain $$x_* \not\in X$$.

Hence $$X$$ cannot be compact if $$X$$ has no largest element even if $$X$$ does have a smallest element.

By analogous reasoning we can also show that if $$X$$ has no smallest element, then $$X$$ cannot be compact, even if $$X$$ does have a largest element.

Am I right?

Now what about the converse? That is, if $$X$$ is a simply ordered set having a smallest element and a largest element, then can we prove that $$X$$ is always compact?

• @JoséCarlosSantos what have you edited in my post? – Saaqib Mahmood Feb 24 at 11:52
• I have added the proof-verification tag. – José Carlos Santos Feb 24 at 11:54
• The empty set might be considered a counterexample. It is a compact space, and it has no largest or smallest element. I don't know, does the definition of an ordered set require the set to me nonempty? – bof Feb 24 at 12:08

On the other hand, $$[-1,1]\setminus\{0\}$$ is not compact with respect to the order topology induced by its usual order, in spite of the fact that it has both a smallest and a largest element.
If $$X$$ is a (LOTS) linearly ordered topological space and $$X$$ is compact, then for any subset of $$A$$, $$\sup A$$ exists in $$X$$. This in fact characterises compact LOTSes. In particular $$\sup X = \max X$$ and and $$\sup \emptyset = \min X$$ exist.
A corollary to the above is that an order complete LOTS $$X$$ is compact iff it has a maximum and a minimum. José's example shows we cannot completeness; endpoints alone are not enough,there shouldn't be any "holes".
As to your proof, if $$X$$ has a minimum $$a$$ and if $$X$$ has no maximum then your family is indeed an open cover without a finite subcover, contradicting the compactness. So for the case at hand this suffices. But I also wanted to point out the general result for LOTSes.
• What if $X=\emptyset$? Does the definition of a LOTS require it to be nonempty? – bof Feb 24 at 12:09