Prove something is an open set

[l888l888l888]

Homework Statement

prove (a,b) x (c,d) is open in R^2

Homework Equations

The Attempt at a Solution

given an x that belongs to (a,b)x(c,d) there exists a delta such that given any y that belongs to R^2 whose d2(x,y)< epsilon, y also belongs to (a,b)x(c,d). Need to prove this.

 

[gb7nash]

Look as R = (a,b)x(c,d) as an open rectangle. Fix y in (a,b)x(c,d). How do you think we can find an open ball(centered at y) that's entirely inside the rectangle?

Hint: Consider the distance of y from a,b,c, and d.

 

[l888l888l888]

to find a ball that is in the rectangle with center y, we can imagine the ball as having radius b-y, y-a, d-y or y-c...

 

[Fredrik]

Are all of those numbers small enough to guarantee that the open ball around y is a subset of the rectangle?

Your suggest in the attempt part of post #1 mentions both a delta and an epsilon. No need for that. Just let y be an arbitrary point in the rectangle and find an r such that B(y,r) (the open ball around y with radius r) is a subset of the rectangle.

 

[Dick]

to find a ball that is in the rectangle with center y, we can imagine the ball as having radius b-y, y-a, d-y or y-c...

You've got the right idea, but if you take a point in (a,b)x(c,d) it's not just a number. It's a pair of numbers. Let's call it (x,y) instead. Now what are the choices of radii, and which radius do you choose?

 

[l888l888l888]

in an answer the first reply... anything bigger than those numbers would imply that the ball would have points inside it but not inside the rectangle... In an answer to the second reply... using (x,y) instead then a possible radius would be b-x, x-a, d-y, or y-c... we could choose b-x as the radius

 

[Dick]

in an answer the first reply... anything bigger than those numbers would imply that the ball would have points inside it but not inside the rectangle... In an answer to the second reply... using (x,y) instead then a possible radius would be b-x, x-a, d-y, or y-c... we could choose b-x as the radius

You wouldn't want to choose b-x as the radius if it were larger than x-a, would you? Wouldn't that make the ball contain points outside of the rectangle, as you said. How about choosing a minimum of those numbers?

 

[l888l888l888]

so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.

 

[l888l888l888]

oh. have we completed the proof?

 

[Dick]

so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.

Well, the ball around (x,y) of radius z would be contained in the open rectangle, right? It's radius is less than or equal to distance to any edge, yes?

 

[gb7nash]

so letting z= min{b-x, x-a, d-y, y-c} now what?. Using this with the info from first replay. we can find a ball with radius z st the ball is contained in the rectangle.

Since your open ball centered at (x,y) with radius z is entirely contained in (a,b)x(c,d), what can you conclude?

 

[l888l888l888]

I have another question about open sets. Is it possible in a metric space containing more than one point to have only the empty set and the the metric space itself as it's open sets? I say no because if you pick a point that point itself with the area around it is a subset of the metric space and u can always find a ball around it with radius small enough such that that ball is in the subset... What do you guys think>

 

[gb7nash]

I have another question about open sets. Is it possible in a metric space containing more than one point to have only the empty set and the the metric space itself as it's open sets? I say no because if you pick a point that point itself with the area around it is a subset of the metric space and u can always find a ball around it with radius small enough such that that ball is in the subset... What do you guys think>

Absolutely right.

For any open set, you can always find a smaller set that's open. Just pick an arbitary point (call it x) in the set. By the definition of open, x is an interior point. Choose a radius that's small enough and create an open ball around x. This open ball is a subset of the entire set and it is not equal to the entire set.