The abnormal Sorgenfrey plane

by math_explorer, Apr 23, 2017, 3:38 AM

aka "I copy my solution to a pset and call it a post to keep my blog alive." To be fair, this is a pretty from-first-principles proof I came up with instead of consulting a book or Google. (Hopefully it's correct!)

Annoying thing: $[a, b]$ and $[a, b)$ always denote closed and half-open intervals, but $(a, b)$ sometimes denotes an open interval and sometimes denotes a point via its two coordinates, usually the latter. I'm sorry.

Definition.

The Sorgenfrey line is the topology on $\mathbb{R}$ generated by the basis of all half-open intervals of the form $[a, b)$ for $a, b \in \mathbb{R}, a < b$ . It is denoted $\mathbb{R}_\ell$ .
The Sorgenfrey plane is $\mathbb{R}_\ell^2$ , i.e. the product of the Sorgenfrey line with itself.

Fact. The Sorgenfrey line is finer than the standard topology on $\mathbb{R}$ (all open sets in the standard topology are also open on the Sorgenfrey line, but not vice versa.) Proof: $U$ is open in the standard topology iff for all $x \in U$ we have $x \in (a, b) \subseteq U$ ; then $x \in [x, b) \subseteq U$ so $U$ is open in the Sorgenfrey line. On the other hand, the half-open intervals like $[a, b)$ are not open in $\mathbb{R}$ .

Fact. Half-open intervals $[a, b)$ are also closed, hence clopen. Proof:

If $x < a$ , then $x$ belongs to the open set $[x, a)$ outside $[a, b)$ .
If $x \geq b$ , then $x$ belongs to the open set $[x, x + 1)$ outside $[a, b)$ .

Theorem. The Sorgenfrey plane is regular (for any point and closed set not containing the point, there exist disjoint open sets around them. This terminology is infuriating but it is what it is, sigh.)

Proof. In $\mathbb{R}_\ell^2$ , let $x$ be a point and $A$ be a closed set not containing $x$ . Then $\mathbb{R}_\ell^2 \setminus A$ is open, so there is a basis element $U = [a, b) \times [c, d)$ of $x$ that is contained in that set, and thus disjoint from $A$ . But $U$ is the product of two closed sets, so it's also closed, so $U$ and its complement separate $x$ from $A$ .

——————

Let the antidiagonal $D$ be the subset/subspace of the Sorgenfrey plane consisting of points of the form $(x, -x)$ :
$D := \{(x, -x) : x \in \mathbb{R}_\ell\} \subset \mathbb{R}_\ell^2.$
Fact. $D$ is a closed subset of the Sorgenfrey plane. Proof: for every point $(x, y)$ not in $D$ :

If $x + y > 0$ , then $(x, y)$ belongs to the open set $[x, x + 1) \times [y, y + 1)$ outside $D$ .
If $x + y < 0$ , then $(x, y)$ belongs to the open set $[x, (x-y)/2) \times [y, (y-x)/2)$ outside $D$ .

Fact. As a subspace of the Sorgenfrey plane, $D$ inherits the discrete topology (every set is open (and closed)). Proof: for every point $(x, -x) \in D$ , the basis element $[x, x + 1) \times [-x, -x + 1)$ only intersects $D$ at $(x, -x)$ , so every one-point subset of $D$ is open in $D$ , so after arbitrary unions, every subset of $D$ is open in $D$ . Taking complements, every subset of $D$ is closed in $D$ .

Corollary. Every subset of $D$ is closed in the Sorgenfrey plane. Proof: They're a closed subset of $D$ , which is a closed subset of the Sorgenfrey plane.

Theorem. The Sorgenfrey plane is not normal (it is not true that, for any two disjoint closed sets, there exist disjoint open sets around them. You thought the word "regular" was overused? Topologists, please.)

Proof. Let $D_0$ and $D_1$ be two sets that partition $D$ , such that their projections onto $\mathbb{R}$ are dense in $\mathbb{R}$ under the usual topology. For example, we could take $D_0 \cap (\mathbb{Q} \times \mathbb{Q})$ and let $D_1 = D \setminus (\mathbb{Q} \times \mathbb{Q})$ . By the corollary, these are both closed in $\mathbb{R}_\ell^2$ . We claim that no open sets can separate $D_0$ and $D_1$ .

Suppose for the sake of contradiction that there are open sets $U_0, U_1$ that separate $D_0$ and $D_1$ . Then for each $(x, -x) \in D$ there exists $w(x), h(x)$ such that $[x, x + w(x)) \times [-x, -x + h(x))$ is in whichever $U_i$ such that $(x, -x) \in D_i$ .

We can pick an arbitrary $(x_0, -x_0) \in D_0$ and $d_0 = w(x_0)$ , and then build an iterative sequence as follows:

$(x_i, -x_i) \in D_{i \bmod 2}$
$x_{i-1} < x_i < x_{i-1} + d_{i-1}$
$0 < d_i < \min(w(x_i), x_{i-1} + d_{i-1} - x_i, d_i/2)$

It is possible to pick $x_i$ satisfying the first two conditions because both sets $D_0, D_1$ are dense in $D$ . Then one can just pick $d_i$ in the range given by the third condition. Note that
$x_i + d_i < x_{i-1} + d_{i-1}$ and
$d_i < d_{i-1}/2,$ so from this definition, we can see these properties:

As $i$ increases, $x_i$ is strictly monotonically increasing.
As $i$ increases, $x_i + d_i$ is strictly monotonically decreasing.
$d_i > 0$ , but $d_i$ converges to zero.

Since $\mathbb{R}$ (under the standard topology) is locally compact, we know that the nested nonempty closed intervals $[x_i, x_i + d_i]$ have nonempty intersection. Let $(c, -c)$ belong to all the intervals. Then $x_i \leq c \leq x_i + d_i$ for all $i$ . In fact, the inequalities can be strict because $x_i < x_{i + 1} \leq c \leq x_{i+1} + d_{i+1} < x_i + d_i$ .

$(c, -c) \in D$ , so it belongs to some $D_j$ . Then since $d_i$ converges to zero, there exists $d_i$ with $i \not\equiv j \bmod 2$ such that $d_i < h(c)$ . But then:

Since $c < x_i + d_i$ we have $c - x_i < d_i < h(c)$ , so $-x_i < -c + h(c)$ , so the point $(c, -x_i)$ is in $[c, c + w(c)) \times [-c, -c + h(c)) \subset U_j$ .
Also since $c < x_i + d_i$ , the point $(c, -x_i)$ is in $[x_i, x_i + d_i) \times [-x_i, -x_i + w(x_i)) \subset U_{1-j}$ .

This contradicts the fact that $U_0, U_1$ are disjoint. Thus they cannot separate $D_0$ and $D_1$ , so the Sorgenfrey plane is not regular. $\square$

topology

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The abnormal Sorgenfrey plane

by math_explorer, Apr 23, 2017, 3:38 AM

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