The Sorgenfrey Line is a topological space whose underlying space is the real line. The topology is generated by the basis of the half open intervals where and are real numbers. For students of topology, the Sorgenfrey Line is a handy example of (1) “Lindelof x Lindelof” does not have to be Lindelof, (2) “normal x normal” does not have to be normal, (3) “paracompact x paracompact” does not have to be paracompact, (4) “perfectly normal x perfectly normal” does not have to be perfectly normal (does not even have to be normal). In other words, these four properties are not preserved by taking Cartesian product. The goal of this note is to prove these and a few other results about the Sorgenfrey Line. In this note, is to denote the Sorgenfrey Line.

We will show these results:

- A
- is Lindelof (thus is normal and paracompact).
- B
- is hereditarily Lindelof.
- C
- Compact subsets of are countable. Thus is an example of a space that is Lindelof and not -compact.
- D
- is not Lindelof.
- E
- is not normal.
- F
- is not paramcompact.
- G
- is not second countable, thus not metrizable.
- H
- is perfectly normal.

In proving C, we will use the following lemma (Lemma 1), which was proved in a previous post. This is a special countable extent property of the real line. Note that a space has extent of cardinality if is the least upper bound on the sizes of all closed and discrete subsets of . In proving **F**, the Jones Lemma will be used. This lemma is essentially saying that the extent of a separable normal space cannot be the cardinality continuum or greater. A space is perfectly normal if is normal and every closed subset is a set (equivalently every open subset is an set). These two lemmas are stated below.

**Lemma 1**- Every uncountable subset of has a two-sided limit point.
**Lemma 2 (Jones’ Lemma)**- If is a separable normal space, then it has no closed and discrete subset of cardinality continuum.

See this post for a proof of Jones’ Lemma.

**Proof of A**. Let be an open cover of consisting of open intervals of the form .

Let . We claim that is a countable set. Suppose that is uncountable. By Lemma 1, there exists a real number that is a two-sided limit point of . This means that for every open interval (open interval in the usual topology on the real line) with , the interval contains points of on the left side of as well as on the right side of .

Choose some such that . Note that is a subset of and so should not contain points of . Because is a two-sided limit point of , will contain points of , a contradiction. So must be countable.

Note that is an open set in the usual topology, which is Lindelof. So we can find countably many such that the union of all such covers . Then find countably many that cover the countably many points in . Combining both sets of , we see that has a countable subcover.

**Proof of B**. Take any uncountable subspace of , we can apply the same proof as in A.

**Proof of C**. Let be a compact subspace that is uncoutable. By Lemma 1, has a two-sided limit point (i.e. it is both a left-sided limit point and a right-sided limit point in the usual topology). Let be a sequence of points in that converges to from the left. Then and form an open cover of that has no finite subcover. Thus all compact subspaces of the Sorgenfrey Line are countable. Furthermore is an example of a Lindelof but not -compact space.

**Proof of D**. If is Lindelof, then it would be a separable normal space and cannot have closed and discrete subset of cardinality continuum or greater (according to Jones’ Lemma). Note that is a closed and discrete subset of , which has cardinality continuum. This shows that cannot be Lindeloff.

**Proof of E and F**. For E, use the same argument as in the proof of D. For F, note that paracompactness implies normality.

**Proof of G**. If is second countable (has a countable base), then it would be a separable metrizable space and would also be a separable metrizable space. But is not even Lindeloff. Thus the Sorgenfrey Line cannot be second countable.

**Proof of H**. Let be an open subset of . Let be the interor of in the usual topology. By a similar argument in the proof of A above, we can show that is countable. Since is an open set in the usual topology, it is an set in the usual topology (and thus in the Sorgenfrey topology). It follows that plus countably many points would form an set. Thus every open subset of the Sorgenfrey line is a -set. Since the Sorgenfrey line is Lindelof, it is normal. Thus the Sorgenfrey line is a normal space in which every open set is a -set (i.e. it is perfectly normal).