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How to prove that every compact subspace of the Sorgenfrey line is countable?

Henno Brandsma
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Paul
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2 Answers2

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Let $C$ be a compact subset of the Sorgenfrey line (so $X = \mathbb{R}$ with a base of open sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $\mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+\frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $\mathbb{R}$.

Suppose that $x_0 < x_1 < x_2 < \ldots $ is a strictly increasing sequence in $C$, and let $c = \sup \{x_n: n =0,1,\ldots \}$, which exists and lies in $C$ by the above remarks. Also let $m = \min(C)$, which also exists by the same.

Then the sets $[c,\rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n \ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$. This contradicts that $C$ is compact.

We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.

And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $\mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $\mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).

Henno Brandsma
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  • why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered? –  Nov 06 '18 at 14:08
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    @dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical. – Henno Brandsma Nov 06 '18 at 14:11
  • Doesn't this actually imply that the compact sets must in fact be finite? – HPP_00 Nov 23 '20 at 11:01
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    @HariPrasadPoilath No, there are countable well-ordered sets and an infinite compact example is $C={0} \cup {\frac1n\mid n =1,2,\ldots}$. – Henno Brandsma Nov 23 '20 at 11:03
  • @HennoBrandsma Thanks. Can all the compact sets of the Sorgenfrey line be classified? That is, a subset is compact if and only if: 1. It is closed and bounded. 2. Does not contain a strictly increasing sequence. – HPP_00 Nov 23 '20 at 15:50
  • @HariPrasadPoilath hes, that’s what I showed. – Henno Brandsma Nov 23 '20 at 16:39
  • @HennoBrandsma I didn't see where the reverse implication was done. If a subset is compact, then yes, it has the above properties. But does having these properties imply that the subset is compact? – HPP_00 Nov 24 '20 at 14:34
  • @HariPrasadPoilath that is true, I think. The argument is a little bit more involved. Maybe the easiest is to show that the set is then closed in the Euclidean topology. – Henno Brandsma Nov 24 '20 at 17:24
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Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.

Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.

Colin McQuillan
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