Prove set is open using open ball

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Homework Help Overview

The problem involves proving that the set O = {(y1, y2) : y1 - y2 > 0} is an open subset of R² in the Euclidean metric. Participants are exploring the properties of this set and its relationship to the boundary defined by the line y1 = y2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to determine an appropriate radius for an open ball and question the implications of points in the set relative to the boundary line. There is also confusion regarding the correct interpretation of inequalities related to the set.

Discussion Status

Some participants have provided suggestions for visualizing the set and calculating distances to the boundary. There is acknowledgment of errors in the original post's definition of the set, and participants are actively questioning assumptions and clarifying definitions.

Contextual Notes

Participants express frustration with the complexity of open set questions and mention that the problem is part of a larger context involving proving the closedness of another set. There is a reference to the complement of the set being a union of open sets, which is part of the ongoing discussion.

amanda_ou812
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Homework Statement


Prove that the set O = {(y1, y2) : y1 - y2 > 0} is an open subset of R2 in the Euclidean metric.


Homework Equations





The Attempt at a Solution


Try as I might I cannot figure out what to take for my r. Any suggestions?
 
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Graph the set! It's bounded by the line y1=y2, right? Find the distance from an arbitrary point to that line.
 
I did graph it. It is the graph of y2 > y1. So on the coordinate plane it would be the dashed line y=x shaded above. I just guessed that my r might be (y1 +y2)/2 but when I try all my inequalities I can't get to x2>x1 (assuming that (x1, x2) is my other point in the ball that I need to show is in O)
 
Wait, but if (y1, y2) is in the open set, the y2-y1>0 which implies that y1-y2<0 so then wouldn't (y1-y2)/sqrt 2 be less than 0? I always have the hardest time with these open set questions
 
amanda_ou812 said:
Wait, but if (y1, y2) is in the open set, the y2-y1>0 which implies that y1-y2<0 so then wouldn't (y1-y2)/sqrt 2 be less than 0? I always have the hardest time with these open set questions

Your original post said y1-y2>0. Seems to have gotten scrambled somewhere along the line. I would have a hard time proving x2<x1 just using basic inequalities. I would just draw the picture. Not good enough for your course??
 
Oh yes, the original post is incorrect. The set is y2-y1>0 or y2>y1. Sorry. No, a picture proof will not due. I am actually doing this as part of a larger problem. Let S be in R2 be the set defined by S = {(a1,a2) : 0 <= a2 <= a1}. Prove that S is a closed subset of R2 in the Euclidean metric. So, I was going about by proving that the complement is open. Well, the complement is the union of two open sets (which means the complement will be open). The first open set is x<0 which I was able to prove easily. The second set is y>x. Which I cannot figure out. Hence, the post. Grrr, the questions are so aggravating. once you think that you understand it, they throw out a question that stumps me.
 

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