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

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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.).

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

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    $\begingroup$ @John But Collin's right. $\endgroup$ – martini Jul 12 '12 at 12:59
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    $\begingroup$ 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 ... $\endgroup$ – martini Jul 12 '12 at 13:06
  • $\begingroup$ @John: see my second hint above. $\endgroup$ – Colin McQuillan Jul 12 '12 at 13:44
  • $\begingroup$ 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. $\endgroup$ – Temitope.A Jul 20 '12 at 13:08
  • $\begingroup$ @Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact. $\endgroup$ – Henno Brandsma Jan 28 at 9:14
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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

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  • $\begingroup$ 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?? $\endgroup$ – noctusraid Feb 10 '16 at 9:27

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