# Showing A Subset of the Product Space ##Y \times Y## is Open

• Bashyboy
In summary: Suppose that there is an element ##e## between ##x## and ##y##. Then the basis elements for the product topology would be ##(a,e)## and ##(e,d)##, which are not open.
Bashyboy

## Homework Statement

Let ##Y## be some order spaced endowed with the order topology, and let ##P = \{(x,y) \in Y \times Y ~|~ x < y \}##. I would like to show that ##P## is open in ##Y \times Y##.

## The Attempt at a Solution

Let us momentarily assume that ##Y## has neither a smallest nor largest element. In this case, the basis elements for the product topology are of the form ##(a,b) \times (c,d)##, where ##(a,b)## is an open interval in ##Y##.

Let ##\langle x,y \rangle## be some point in ##P \subseteq Y \times Y##. ##x,y \in Y##, there exists basis elements ##(a,b)## and ##(c,d)## such that ##x \in (a,b)## and ##y \in (c,d)##. Now, if there is an element ##e## between ##x## and ##y##, we can "readjust" the basis elements to obtain ##(a,e)## and ##(e,d)##, and clearly every element ##(a,e)## is smaller than every element in ##(e,d)##. Hence, ##\langle x,y \rangle \in (a,e) \times (e,d) \subseteq P##.

Now, suppose that there is no element between them. Then ##(a,y)## will contain ##x## and ##(x,d)## will contain ##y##. Again, it isn't difficult to see that every element the latter set is smaller than every element in the former set. In either case, we can conclude that ##P## is open.

Does this sound right? I suspect that the when ##Y## has a smallest or largest element the proof will be very similar. If so, I will omit it.

Bashyboy said:
suppose that there is no element between them.
I believe your argument for this case also works when there is an intervening element.
Bashyboy said:
when Y has a smallest or largest element the proof will be very similar
From a quick reading of the subject (never met it before) you can suppose a basis consisting of open "rays" (-∞, x), (x, ∞). I assume the infinities represent min and max elements if they exist. So you might be able to write one argument that covers all cases.

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haruspex said:
I believe your argument for this case also works when there is an intervening element.

Are you sure? I need an open set about ##U_x## about ##x## and an open set ##U_y## about ##y## such that everything in ##U_x## is strictly smaller than everything in ##U_y##. If I choose the sets I did above and there is an intervening element won't we have ##e < e##, which is a problem. I am sorry; I am not thinking clearly at the moment, and its pretty noisy where I am.

haruspex said:
From a quick reading of the subject (never met it before) you can suppose a basis consisting of open "rays" (-∞, x), (x, ∞). I assume the infinities represent min and max elements if they exist. So you might be able to write one argument that covers all cases.

In my book, basis elements for the order topology are defined as ##(a,b)##, ##[a_0,b)##, where ##a_0## is the smallest element (if it exists), and ##(a,b_0]##, where ##b_0## is the largest element (if it exists). But I think you are right that we can knock out all cases by just looking at ##(-\infty, a)## and ##(a,\infty)## and letting the infinities represent the largest/smallest elements or finite elements. , since the mechanics of the proof will be the same in each case. I would like verification, of course.

Bashyboy said:
If I choose the sets I did above and there is an intervening element won't we have e<ee<ee < e, which is a problem.
You are right, and thinking more clearly than I am.

Okay, I think I may have made progress towards slightly more understanding. In my textbook, it is shown that elements of the form ##(a, \infty)## and ##(- \infty, a)## form a subbasis for the order topology, and a subbasis generates the topology through arbitrary unions of finite intersections of subbasis elements; i..e., letting ##\mathcal{S}## denote the set of all open rays, then a open set will be of the form ##\bigcup_{i \in I} S_i##, where ##S_i## is a finite intersection of elements in ##\mathcal{S}##. So you are right that we can just look at these open rays.

This actually helps with another problem I am working on.

## 1. What is a subset of the product space ##Y \times Y##?

A subset of the product space ##Y \times Y## is a collection of elements that are contained within the larger set of all possible pairs of elements from the set Y.

## 2. Why is it important to show that a subset of the product space ##Y \times Y## is open?

Showing that a subset of the product space ##Y \times Y## is open allows us to prove that the subset satisfies the properties of openness, which are essential for many mathematical proofs and applications.

## 3. What is the definition of an open set in a product space?

An open set in a product space is a subset that contains all of its interior points, meaning that for every point in the subset, there exists a neighborhood around that point that is also contained within the subset.

## 4. How can we prove that a subset of the product space ##Y \times Y## is open?

There are a few different methods for proving that a subset of the product space ##Y \times Y## is open. One way is to show that the subset can be written as a union of open sets, or that it satisfies the definition of openness by having all of its interior points contained within the subset.

## 5. Can you provide an example of a subset of the product space ##Y \times Y## that is open?

One example of a subset of the product space ##Y \times Y## that is open is the set of all points that lie on or inside a circle with a given radius in the Cartesian plane. This subset contains all of its interior points and can be written as a union of open intervals, making it an open set in the product space.

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