- #1

Bashyboy

- 1,421

- 5

## 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.