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

1. Dec 2, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
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 $Y \times Y$.

2. Relevant equations

3. 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.

2. Dec 3, 2016

### haruspex

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

Last edited: Dec 3, 2016
3. Dec 3, 2016

### Bashyboy

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.

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.

4. Dec 3, 2016

### haruspex

You are right, and thinking more clearly than I am.

5. Dec 4, 2016

### Bashyboy

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.