How to prove that compact subspaces of the Sorgenfrey line are countable?

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

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.).

• why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered? – creepyrodent Nov 6 '18 at 14:08
• @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 6 '18 at 14:11

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.

• @John But Collin's right. – martini Jul 12 '12 at 12:59
• You should try it yourself, for starters. And in the classical topology of $\mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ... – martini Jul 12 '12 at 13:06
• @John: see my second hint above. – Colin McQuillan Jul 12 '12 at 13:44
• The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret. – Temitope.A Jul 20 '12 at 13:08
• @Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact. – Henno Brandsma Jan 28 at 9:14

not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R

• this doesn't make any sense. [0,1] is compact in $\mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway?? – noctusraid Feb 10 '16 at 9:27