3
$\begingroup$

This question already has an answer here:

The Heine–Borel theorem perfectly characterizes the compact subsets of the real line $\mathbb{R}$ (with the usual metric/order topology):

Heine–Borel Theorem. A subset $A \subseteq \mathbb R$ is compact if and only if it is closed and bounded.

What characterisations are there for the compact subsets of the Sorgenfrey line, $\mathbb{S}$?

Since the topology on $\mathbb S$ is finer than the topology on $\mathbb R$, every compact subset of $\mathbb{S}$ is also a comapct subset of $\mathbb{R}$, and so they must be closed (as subsets of $\mathbb{R}$) and bounded. But more is needed: $[0,1]$ is not compact, since it has an infinite disjoint open cover: $$ [0,\tfrac 12) , [\tfrac 12, \tfrac 34 ) , [ \tfrac 34 , \tfrac 78 ) , \ldots , [ \tfrac{2^n-1}{2^n}, \tfrac{2^{n+1}-1}{2^{n-1}}), \ldots , [1 , 2 ).$$

$\endgroup$

marked as duplicate by tomasz, Davide Giraudo, Jyrki Lahtonen, TrueDefault, Claude Leibovici Sep 10 '14 at 8:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

In this answer I sketch the answer: a subset $K$ of the Sorgenfrey line is compact, iff it is closed, bounded and is well-ordered under $>$ (i.e. every non-empty subset of $K$ has a maximum). In particular, all such subsets are at most countable.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.