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.


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?

  • $\begingroup$ @JoséCarlosSantos what have you edited in my post? $\endgroup$ – Saaqib Mahmood Feb 24 at 11:52
  • $\begingroup$ I have added the proof-verification tag. $\endgroup$ – José Carlos Santos Feb 24 at 11:54
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    $\begingroup$ 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? $\endgroup$ – bof Feb 24 at 12:08

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.


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.

  • $\begingroup$ What if $X=\emptyset$? Does the definition of a LOTS require it to be nonempty? $\endgroup$ – bof Feb 24 at 12:09
  • $\begingroup$ @bof of course, all my spaces are non-empty. $\endgroup$ – Henno Brandsma Feb 24 at 12:29

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