A Note On The Sorgenfrey Line

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 [a, b) where a and b 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, S is to denote the Sorgenfrey Line.

We will show these results:

S is Lindelof (thus is normal and paracompact).
S is hereditarily Lindelof.
Compact subsets of S are countable. Thus S is an example of a space that is Lindelof and not \sigma-compact.
S \times S is not Lindelof.
S \times S is not normal.
S \times S is not paramcompact.
S is not second countable, thus not metrizable.
S 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 X has extent of cardinality \mathcal{K} if \mathcal{K} is the least upper bound on the sizes of all closed and discrete subsets of X. 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 X  is perfectly normal if X is normal and every closed subset is a G_{\delta} set (equivalently every open subset is an F_{\sigma} set). These two lemmas are stated below.

    Lemma 1
    Every uncountable subset of \mathbb{R} has a two-sided limit point.
    Lemma 2 (Jones’ Lemma)
    If X 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 \mathcal{A} be an open cover of S consisting of open intervals of the form [a, b).

Let T=\cup\lbrace{(a, b): [a, b)\epsilon\mathcal{A}}\rbrace. We claim that U=S - T is a countable set. Suppose that U is uncountable. By Lemma 1, there exists a real number p that is a two-sided limit point of U. This means that for every open interval (s,t) (open interval in the usual topology on the real line) with p \in (s,t), the interval (s,t) contains points of U on the left side of p as well as on the right side of p.

Choose some [a,b) \in \mathcal{A} such that p \in [a,b). Note that (a,b) is a subset of T and so should not contain points of U=S-T. Because p is a two-sided limit point of U, (a,b) will contain points of U, a contradiction. So U must be countable.

Note that T is an open set in the usual topology, which is Lindelof. So we can find countably many [c,d) \in \mathcal{A} such that the union of all such (c,d) covers T. Then find countably many [c,d) \in \mathcal{A} that cover the countably many points in U=S-T. Combining both sets of [c,d), we see that \mathcal{A} has a countable subcover. \blacksquare

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

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

Proof of D. If S \times S 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 D=\lbrace{(x,-x): x\epsilon{S}}\rbrace is a closed and discrete subset of S \times S, which has cardinality continuum. This shows that S \times S cannot be Lindeloff. \blacksquare

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

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

Proof of H. Let W be an open subset of S. Let O be the interor of W in the usual topology. By a similar argument in the proof of A above, we can show that W-O is countable. Since O is an open set in the usual topology, it is an F_{\sigma} set in the usual topology (and thus in the Sorgenfrey topology). It follows that O plus countably many points would form an F_{\sigma} set. Thus every open subset of the Sorgenfrey line S is a G_\delta-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 G_\delta-set (i.e. it is perfectly normal). \blacksquare

13 thoughts on “A Note On The Sorgenfrey Line

  1. Hi Dan, i was searching something about the paracompactness of the Sorgenfrey Line and I found this fantastic blog! I have to praise you for your work and I assure you I will follow it since now!
    However, I had some doubts reading this article. First, you said: “S is not second countable, thus not metrizable”, but I thought that \mathbb{R} with the “trivial” distance d:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R},\ s.t.\ \forall\ x,y\in \mathbb{R}: d(x,y)= \begin{cases}   0, & \mbox{if } x \neq y\\   1, & \mbox{otherwise} \end{cases} was an example of metric space which does not have to be second-countable.
    The other doubt was: how can you deduce that the Sorgenfrey Line is paracompact from the fact that it is Lindeloff?
    Sorry for the triviality,

    • Gianluca, your example of the trivial metric on the real line only shows that there is a non-second countable metrizable topology that can be defined on the real line. However, this “trivial” metric topology has nothing in common with the Sorgenfrey topology. I think your definition of the trivial metric is not what you intended. You probably meant that d(x,y)=0 if x=y and d(x,y)=1 if x not equal to y. Even with this correction, your example does not cast any doubt on the statement that the Sorgenfrey line is not second countable and thus not metrizable. On your second point, it is a well known theorem in general topology that any regular Lindelof space is paracompact. A good reference is Engelking (theorem 3.8.11 and theorem 5.1.2). Thanks for your comment.

  2. Sorry, the definition of the metric I exposed in the first comment was obviously wrong, I meant exactly what you wrote. But I still don’t understand this sentence: “is not second countable and thus not metrizable”. Perhaps I have some problems with English (I’m italian), but for me this statement means “if a topological space is not second-countable, then it is not metrizable”, so “if a topological space is metrizable, then it is second-countable”, and this one does not have to be true.
    Thanks for your reply.

    • Giaorl, I now understand your question. In general, metrizable does not imply second countable (the discrete topology you mentioned is a handy example). However, for separable spaces, metrizable does imply second countable. The Sorgenfrey line is both separable and Lindelof. Thus if S is metrizable, it would be a separable metric space and would have a countable space. Consequently S x S would be metrizable too. But we know that S x S is not even Lindelof. The point G in the post should restated as “S is not metrizable” rather than “S is not second countable, thus not metrizable”. I now see that taken by itself (without the right context), it could be misleading.

      • Well, now I understand! I’m studying for my first exam in General Topology, so i know only basic theorems. Thanks!

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