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

Homework Equations

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.

 

[haruspex]

suppose that there is no element between them.

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

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.

 

[Bashyboy]

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.

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.

 

[haruspex]

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.

 

[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.