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### Homework Help 2003

From: Henno Brandsma
Date: Dec 14, 2003
Subject: Re: Baireness of Sorgenfrey line, more details (and more accurate)

To prove that S (the Sorgenfrey line, also known as the lower limit topology
on R, or the "right half open interval topology" in Counterexamples in Topology)
is a Baire space (for every countable family of open dense subsets O_n of S
the intersection is dense) we will use its special relationship to the
usual topology on R. Denoting the topology on R by T(R) and the Sorgenfrey topology on R
(which has a base of subsets of the form [a,b), where a < b) by T(S), then we have that

1) T(R) \subset T(S): if (a,b) is a basic open subset of T(R), then for each
x in (a,b), the subset [x,b) is a subset of (a,b), and this shows that each x in (a,b)
is an interior point of (a,b) in T(S). So (a,b) is Sorgenfrey-open, and all
open sets of T(R) are unions of open intervals, so all open subsets in T(R) are
in T(S) as well.

2) Every non-empty open subset of T(S) has a non-empty interior in the T(R)-topology.
For this it suffices to consider sets of the form [a,b) (a < b), and these have interior (a,b)
in T(R), and this is non-empty.

\footnote{Technicality: we could say that T_1\{\emptyset} is coinitial in T_2\{\emptyset} instead
of 1)+2)}

Now we prove two simple but useful lemma's:

Lemma 1: if a set X has two topologies T_1 \subset T_2 such that each non-empty
A in T_2 has non-empty interior in T_1, and O is open and dense in (X,T_2) then
Int(O) (interior in T_1) is also dense (and of course open) in T_1.

Proof: Let U be open and non-empty in (X,T_1). Then U is also open in T_2
and so it intersects O (this uses that O is dense in T_2). The intersection
U /\ O is open in T_2, and non-empty, so it has non-empty interior in T_1.
But Int(O/\U) \subset Int(O) and Int(O/\U) \subset Int(U)=U (all Int in T_1)
and so we see that U intersects Int(O) in at least the non-empty set Int(O/\U).
We have shown now that every non-empty open U intersects Int(O) and hence the latter
set is dense, as claimed.

Lemma 2: let X, T_1 and T_2 be as before. If D \subset X is dense in T_1 then
it is dense in T_2.

Proof: Let U be a non-empty open subset of (X,T_2). Then Int(U) (Int in T_1) is non-empty
and open in T_1, so it intersects the dense subset D of T_1.
But if D intersects Int(U), it certainly intersects U, so D intersects every
non-empty open set of T_2 and we are done.

Now we are ready to prove

Theorem 1: If X, T_1 and T_2 are as in the lemma's, then (X,T_1) is Baire implies that
(X,T_2) is Baire.

Proof: Suppose (X,T_1) is Baire and let O_n (n in N) be open and dense subsets of (X,T_2).
By Lemma 1, Int(O_n) (in (X,T_1)) are open and dense, and by Baireness of (X,T_1) we conclude
that /\Int(O_n) is dense in (X,T_1). But then the larger subset D = /\_n O_n is also
dense in (X,T_1), and by Lemma 2 we see that D is also dense in (X,T_2), which is
exactly what we needed to show. This concludes the proof.

Corollary: the Sorgenfrey line is Baire.
Proof: the remarks in the beginning show that X = R and T_1 = T(R) and T_2 = T(S)
satisfy the conditions in the lemma's. Moreover, (R,T(R)) is a Baire space
because it is locally compact, or alternatively because it is a complete metric space.
The theorem now does the rest.

---

Henno Brandsma