How to prove that every compact subspace of the Sorgenfrey line is countable?
How to prove that compact subspaces of the Sorgenfrey line are countable?
3 Answers
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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1@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. – 2018-11-06
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.
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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0this 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?? – 2016-02-10