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Maybe my misunderstanding of topology and compact sets is seriously flawed, but as far as I can tell I can't see why the "Sorgenfrey Line" is not $\sigma$-compact.

A topological space is $\sigma$-compact if it is the countable union of compact subsets. The Sorgenfrey Line is simply the real line endowed with the topology of half-open intervals. As stated on Wikipedia, the sets $(-\infty, a)$ and $[b, \infty)$ are both open, implying $[a, b)$ is clopen, and hence compact for finite $a, b$ (closed and bounded, unless Heine-Borel doesn't hold in this space for some reason).

Then we can write $\mathbb{R} = \bigcup_{n\in\mathbb{Z}}[n-\epsilon,n+1)$ for some $\epsilon>0$, so the Sorgenfrey line is clearly $\sigma$-compact.

What's the flaw in the logic? Is it the assumption that Heine-Borel holds in this space, and so the subsets with finite open subcovers aren't exactly the closed, bounded sets?

user3002473
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  • Here is a proof that a compact sunset of the Sorgenfrey line is at most countable. – saulspatz May 07 '18 at 21:26
  • The Heine-Borel theorem is a statement about metric spaces. So it doesn't apply to the Sorgenfrey line. – Rob Arthan May 07 '18 at 21:28
  • As for your question: $[0,1)=\bigcup_{n\in\mathbb{Z}}[0,1-\frac{1}{n})$, which has no finite subcover. Indeed the problem is that Heine-Borel doesn't hold in this space – Notone May 07 '18 at 21:28
  • The Heine-Borel theorem is little more than the result that [0, 1] is compact in the ordinary real line. This is not the ordinary real line, so there's no reason to expect it to hold here, any more than one would expect it to hold in infinite-dimensional spaces, arbitrary metric spaces, etc. – anomaly May 07 '18 at 21:55

1 Answers1

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Heine-Borel is specific to the Euclidean topology, and need not hold in a general space (where "bounded" makes no sense, necessarily), and it does not hold for the Sorgenfrey line: $[0,1]$ can be covered by the open sets $[0,1-\frac{1}{n})$, $n \in \mathbb{N}$ and $[1,2)$ and this cover has no finite subcover. So $[0,1]$ is not compact.

In fact, I showed here that a compact subset of the Sorgenfrey line is at most countable. So a $\sigma$-compact subset is also at most countable.

As a bonus an indirect proof: suppose $S$ were $\sigma$-compact. Then $S \times S$ would be $\sigma$-compact (hence Lindelöf) as well, and $T_3$, and a $T_3$ Lindelöf space is normal while $S \times S$ is not normal.

Or, $D':= \{(x,-x) : x \in S\} \subset S \times S$ is closed, uncountable and discrete and so $S \times S$ cannot be $\sigma$-compact, as otherwise $D'$ would be too, which it is not: the only $\sigma$-compact discrete spaces are the countable ones.

Henno Brandsma
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  • That proof that the compact subsets of the Sorgenfrey Line are at most countable is absolutely bizarre. The properties of these objects are so alien. Thanks for the explanation. – user3002473 May 07 '18 at 21:35
  • @user3002473 Glad I could be of help. $\mathbb{S}$ is one of my favourite spaces.. – Henno Brandsma May 07 '18 at 21:36