16

It is very easy to prove that the Sorgenfrey line is completely regular:

To separate a point $x$ from a closed set $F$, note that there is an interval $[x,y)$ disjoint from $F$ and observe that the characteristic function of $[x,y)$ is continuous because the half-open intervals generating the topology are clopen.

The proof of normality seems considerably harder. The argument I found in several books proceeds in two steps:

  1. Show that the Sorgenfrey line is Lindelöf.

  2. Show that a regular Lindelöf space is normal.

Both these steps are much harder than the above argument. Putting those two arguments together, one obtains something like the argument for point H in Dan Ma's a note on the sorgenfrey line.

Is there a shortcut that is simpler than putting these two arguments together?

Martin Sleziak
  • 56,060
janos s.
  • 161

2 Answers2

31

Suppose $A,B$ are disjoint closed sets in the Sorgenfrey line. We have to find open sets $U,V$ such that $A\subseteq U,\ B\subseteq V,$ and $U\cap V=\emptyset.$

For each $a\in A$ choose $a'\gt a,$ $\ [a,a')\cap B=\emptyset$; then $U=\bigcup\limits_{a\in A}[a,a')$ is an open set containing $A.$

For each $b\in B$ choose $b'\gt b,$ $\ [b,b')\cap A=\emptyset$; then $V=\bigcup_\limits{b\in B}[b,b')$ is an open set containing $B.$

To show that $U\cap V=\emptyset,$ it suffices to show that $[a,a')\cap[b,b')=\emptyset$ for all $a\in A,\ b\in B.$ Suppose $a\in A,\ b\in B,$ and assume without loss of generality that $a\lt b.$ Since $[a,a')\cap B=\emptyset,$ it follows that $b\ge a'$ and so $[a,a')\cap[b,b')=\emptyset.$

bof
  • 82,298
  • 2
    Why can you choose the $a's$ and $b's$ like that? – Travis Oct 24 '20 at 09:37
  • 6
    Given $a\in A$, why can I choose $a'\gt a$ with $[a,a')\cap B=\emptyset$? Well, $A$ and $B$ are disjoint, and $a\in A$, so $a\notin B$. Since $B$ is closed and $a\notin B$, some neighborhood of $a$ is disjoint from $B$. In the Sorgenfrey topology, the neighborhoods of $a$ have a base consisting of sets of the form $[a,a')$. Okay? – bof Oct 24 '20 at 09:45
10

That argument is as easy as any that I know, and its second part is an important result in its own right. An alternative is to prove that every linearly ordered topological space is hereditarily normal and then show that the Sorgenfrey line is a subspace of a linearly ordered space. The second step is easy. Let $X=\Bbb R\times\{0,1\}$ ordered lexicographically: $\langle x,i\rangle\preceq\langle y,j\rangle$ if and only if $x<y$, or $x=y$ and $i\le j$. Give $X$ the order topology generated by $\preceq$, and let $Y$ be the subspace $\Bbb R\times\{1\}$ with the topology that it inherits from $X$. If $\Bbb S$ denotes the Sorgenfrey line, the map

$$f:\Bbb S\to Y:x\mapsto\langle x,1\rangle$$

is easily seen to be a homeomorphism. $Y$ is normal as a subspace of the linearly ordered space $X$, so $\Bbb S$ is normal.

Community
  • 1
Brian M. Scott
  • 631,399