# Show the openness of a set

1. Sep 25, 2007

### alligatorman

I want to show that the set $$S=\{(x,y)\in \mathbb{R}^2 | xy\neq1\}$$ is open.

I'm having trouble forming the open ball contained in S centered at each point in S. The idea I have is:

Let $$q\in S$$. Then select an open ball $$B_r(q), r\in\mathbb{R}$$. Let $$P=\{(x,y)\in B_r(q)|xy=1\}$$. If P is empty, we are done. If P is not empty, then create an open ball $$B_m(q),$$ where $$m=\min\{d(q-c)|c\in B_r(q)\}$$ (d is the distance function). Then we are done.

However, I feel like this isn't sufficient because it has not been shown whether $$\{d(q-c)|c\in B_r(q)\}$$ actually does have a minimum, or if it contains elements tending to 0. Am I just missing a property of the reals somewhere? I hope so. Any help would be appreciated.

2. Sep 25, 2007

### AiRAVATA

Don't you think is easier to prove that $S^c$ is closed?

3. Sep 25, 2007

### alligatorman

In metric spaces, a set is defined to be closed if it's complement is open.

4. Sep 25, 2007

### Hurkyl

Staff Emeritus
Is there any way you can leverage your knowledge of the topology of R to help?

5. Sep 25, 2007

### AiRAVATA

True, so why don't you prove that the complement of S is closed?

6. Sep 25, 2007

### alligatorman

Are you suggesting to prove by contradiction that it is closed? I'm not sure how to go about that because the only definition of closed I have is that it's complement is open.

7. Sep 25, 2007

### HallsofIvy

If x is in the set {(x,y)| xy is not equal to 1} then obviously either xy< 1 or xy>1. Why not look at the two cases separately?

8. Sep 25, 2007

### alligatorman

I could, but how would proving that the set is open be any easier when the cases are split up? I would still have to construct a ball such for each (x,y), no matter how close I make to the boundary.

I'm looking at some theorems, and I have that any subset S of a metric space is bounded if it can be contained in a ball. I can easily contain the set of the d(s) in a ball, which then means it's bounded from above and below. So then I can create the open ball with radius from d(q, lower bound)

9. Sep 25, 2007

### Hurkyl

Staff Emeritus
Just checking: have you yet learned what continuous means?

10. Sep 25, 2007

### alligatorman

I have not. (Not formally. I can't use it in the proof)

11. Sep 25, 2007

### Hurkyl

Staff Emeritus
Ah. So you could not use the fact that multiplication is continuous, and the set of real numbers not equal to 1 is open. Oh well; that's the easy way to do this problem.

12. Sep 25, 2007

### morphism

Suppose it's not open. Then there exists a pair (x,y) in S such that for all r>0, Br $\cap$ Sc is nonempty. In particular, we can do this for all r=1/n (n=1,2,3,...), and thus we obtain for each n a pair (an, bn) satisfying two properties:
(i) an * bn = 1, and
(ii) d((an, bn), (x,y)) < 1/n.

Can you show that this gives us a contradiction? (Hint: think sequences.)

Last edited: Sep 25, 2007