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Sorgenfrey line
The Sorgenfrey line is a nonstandard topology on the real line . Its topology is defined by the following base of half open intervals
Another name is lower limit topology, since a sequence converges only if it converges in the standard topology and its limit is a limit from above (which, in this case, means that at most finitely many points of the sequence lie below the limit). For example, the sequence converges to , while does not.
This topology is finer than the standard topology on . The Sorgenfrey line is first countable and separable, but is not second countable. It is therefore not metrizable.
References
- 1 R. H. Sorgenfrey, On the topological product of paracompact spaces, Bulletin of the American Mathematical Society 53 (1947) 631–632. (This paper is available on-line from Project Euclid.)
Defines:
lower limit topology
Synonym:
Sorgenfrey topology
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Example
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Reference
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Mathematics Subject Classification
55-00 General reference works (handbooks, dictionaries, bibliographies, etc.)54-00 General reference works (handbooks, dictionaries, bibliographies, etc.)
22-00 General reference works (handbooks, dictionaries, bibliographies, etc.)
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new question: pure subgroups by lvoyster
new correction: Typo in M\"obius function? by Aleph Zero
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Corrections
parenthesis by dublisk ✘
lower limit topology by AxelBoldt ✓
"limit from above" by AxelBoldt ✓
Original reference by rspuzio ✓
wording by mps ✓
lower limit topology by AxelBoldt ✓
"limit from above" by AxelBoldt ✓
Original reference by rspuzio ✓
wording by mps ✓



Comments
is the sorgenfrey line compact?
Is the sorengfrey line compact?
Re: is the sorgenfrey line compact?
No. The simplest way to see it is to take the cover
{ [n,n+1[ | n integer }. All the sets are open and if you remove
any one of them, it is no longer a cover. Hence it cannot have
a finite subcover and so cannot be compact.
Is The Sorgenfrey topology a Baire space?
I know that the Sorgenfrey topology is totally disconnected, but I cant seem to prove that this implies it is a baire space. I have given up trying to prove directly that it is a baire space, I couldn't get anywhere with that.
Re: Is The Sorgenfrey topology a Baire space?
Not every totally disconnected space is a Baire space: take the rationals, for example.
But the Sorgenfrey line is indeed a Baire space, and the proof is the same as for the (Euclidean) real line.
compact subsets of the Sorgenfrey line
Why any compact subset of sorgenfrey line must be a countable set?