Prove Open Set: S = {(x,y) in $\mathbb{R}^2$ | xy > 1}

[mateomy]

$$Let\,S=\,\{(x,y)\,\in\,\mathbb{R}^2\,|\,xy\,>\,1\}$$
Show S is open.

I've been reading everything I can get my hands on to get a handle on these sort of proofs. I think I have the definitions of everything down solid but my problem seems to be proving things for myself as I think is true for almost everyone who is first confronted with this stuff. This is what I've done so far...

Let xy be defined as a point q. We define a point z $\in$ D(w,r), where $r=\frac{1}{2}d(w,q)$.
Further;
d(w,q) $\leq$ d(q,z) + d(z,w).
Hence, d(q,z) $\geq$ d(w,q) - d(z,w)


I don't know where to move from here. I'm not even sure if I'm on the right track. I know I can substitute d(w,q) as 2r but I don't know if that will help me at all. Also (for clarification) 'D(w,r)' is my 'ball' in $\mathbb{R}^2$ centered at 'w' with a radius 'r'.

Thanks for the help.

 

[SammyS]

$$Let\,S=\,\{(x,y)\,\in\,\mathbb{R}^2\,|\,xy\,>\,1\}$$
Show S is open.

I've been reading everything I can get my hands on to get a handle on these sort of proofs. I think I have the definitions of everything down solid but my problem seems to be proving things for myself as I think is true for almost everyone who is first confronted with this stuff. This is what I've done so far...

Let xy be defined as a point q. We define a point z $\in$ D(w,r), where $r=\frac{1}{2}d(w,q)$.
Further;
d(w,q) $\leq$ d(q,z) + d(z,w).
Hence, d(q,z) $\geq$ d(w,q) - d(z,w)

I don't know where to move from here. I'm not even sure if I'm on the right track. I know I can substitute d(w,q) as 2r but I don't know if that will help me at all. Also (for clarification) 'D(w,r)' is my 'ball' in $\mathbb{R}^2$ centered at 'w' with a radius 'r'.

Thanks for the help.

What's the definition of an open set?

 

[mateomy]

Informally...an open set is any set whose boundary points are not included within that set. Also, if a point within the set was to have an epsilon 'ball' draw around it of an arbitrary radius any point within that ball would also be located within the set.

 

[HallsofIvy]

The first thing you should do is draw the set. What is the boundary of that set? Then let (x, y) be a point in the set and show that is it NOT in that boundary. Use your picture to decide what is the shortest distance from (x, y) to the boundary.

mateomy, "an open set is any set whose boundary points are not included within that set" is not necessarily "informal". Some textbooks use that as the definition of "open sets" (in a metric space, of course) after defining a boundary point, p, of set A to be a point such that every ball centered on p contains some points in A and some points not in A.

 

[genericusrnme]

Informally...an open set is any set whose boundary points are not included within that set. Also, if a point within the set was to have an epsilon 'ball' draw around it of an arbitrary radius any point within that ball would also be located within the set.

This isn't the most beginner friendly definition to go by.
HallsofIvy's definition is much nicer, it gives you a better idea of an open set as a set where you've always got a little freedom to move about.

 

[mateomy]

I've drawn a diagram, that's how I arrived at my triangle inequality. I was thinking about doing something along these lines:


$$r\,=\,\frac{1}{2}d(w,q)$$

So we can rewrite that as 2r, and because we know that d(w,z) is less than our 'ball' radius we can rewrite the entire inequality as

$$d(q,z)\,\geq\,d(w,q)\,-\,d(z,w)\,>\,2r\,-\,r$$

And since we know the r is never zero d(q,z) will always be greater than zero and hence, our 'ball' will always remain within the set.

Does that work? (I'd clean it up a little obviouslly)

Thanks again, everyone.