## by **math_explorer**, Apr 23, 2017, 11:38 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: and always denote closed and half-open intervals, but sometimes denotes an open interval and sometimes denotes a point via its two coordinates, usually the latter. I'm sorry.

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Let the antidiagonal be the subset/subspace of the Sorgenfrey plane consisting of points of the form :

Suppose for the sake of contradiction that there are open sets that separate and . Then for each there exists such that is in whichever such that .

We can pick an arbitrary and , and then build an iterative sequence as follows:

It is possible to pick satisfying the first two conditions because both sets are dense in . Then one can just pick in the range given by the third condition. Note that

and

so from this definition, we can see these properties:

Since (under the standard topology) is locally compact, we know that the nested nonempty closed intervals have nonempty intersection. Let belong to all the intervals. Then for all . In fact, the inequalities can be strict because .

, so it belongs to some . Then since converges to zero, there exists with such that

This contradicts the fact that are disjoint. Thus they cannot separate and , so the Sorgenfrey plane is not regular.

Annoying thing: and always denote closed and half-open intervals, but 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 generated by the basis of all half-open intervals of the form for . It is denoted . - The
*Sorgenfrey plane*is , i.e. the product of the Sorgenfrey line with itself.

**Fact.**The Sorgenfrey line is finer than the standard topology on (all open sets in the standard topology are also open on the Sorgenfrey line, but not vice versa.) Proof: is open in the standard topology iff for all we have ; then so is open in the Sorgenfrey line. On the other hand, the half-open intervals like are not open in .**Fact.**Half-open intervals are also closed, hence clopen. Proof:- If , then belongs to the open set outside .
- If , then belongs to the open set outside .

**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.)Dress Sleeve Summer Women Shoulder Basic Short line A Off Paneled Solid Daily

*Proof.*In , let be a point and be a closed set not containing . Then is open, so there is a basis element of that is contained in that set, and thus disjoint from . But is the product of two closed sets, so it's also closed, so and its complement separate from .——————

Let the antidiagonal be the subset/subspace of the Sorgenfrey plane consisting of points of the form :

**Fact.**is a closed subset of the Sorgenfrey plane. Proof: for every point not in :- If , then belongs to the open set outside .
- If , then belongs to the open set outside .

**Fact.**As a subspace of the Sorgenfrey plane, inherits the discrete topology (every set is open (and closed)). Proof: for every point , the basis element only intersects at , so every one-point subset of is open in , so after arbitrary unions, every subset of is open in . Taking complements, every subset of is closed in .**Corollary.**Every subset of is closed in the Sorgenfrey plane. Proof: They're a closed subset of , 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 and be two sets that partition , such that their projections onto are dense in under the usual topology. For example, we could take and let . By the corollary, these are both closed in . We claim that no open sets can separate and .Suppose for the sake of contradiction that there are open sets that separate and . Then for each there exists such that is in whichever such that .

We can pick an arbitrary and , and then build an iterative sequence as follows:

It is possible to pick satisfying the first two conditions because both sets are dense in . Then one can just pick in the range given by the third condition. Note that

and

so from this definition, we can see these properties:

- As increases, is strictly monotonically increasing.
- As increases, is strictly monotonically decreasing.
- , but converges to zero.

Since (under the standard topology) is locally compact, we know that the nested nonempty closed intervals have nonempty intersection. Let belong to all the intervals. Then for all . In fact, the inequalities can be strict because .

, so it belongs to some . Then since converges to zero, there exists with such that

**Off Basic Paneled Dress line A Solid Short Daily Women Summer Shoulder Sleeve**. But then:- Since we have , so , so the point is in .
- Also since , the point is in .

This contradicts the fact that are disjoint. Thus they cannot separate and , so the Sorgenfrey plane is not regular.