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?
Best Answer
Your proof is correct.
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.