Ordinary Differential Equations by Tenenbaum and Pollard

  • #1
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I am having a hard time understanding the conditions that set a plane to be called a region.

According to definition 2.68, a set in the plane is called a region if it satisfies the following two conditions (p. 14):

1. "Each point of the set is the center of a circle whose entire interior consists of points of the set."

2. "Every two points of the set can be joined by a curve which consists entirely of points of the set."

I am having a hard time understanding the first condition. Can anyone provide an illustration in order to decipher the first condition meaning?

Thank you for help.
 

Answers and Replies

  • #2
RUber
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I think the counterpoint to 1 would be a boundary point. On the boundary, any circle you draw ( no matter how small ) will include some points from inside the domain and some outside.
Condition 1 indicates that the region is open. An example might be R2 with 0< x, y< 1. For any point near (0,0) or (1,1) there will exist a value r small enough that you can draw a circle around the point with only members of the region inside.
 
  • #3
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I think the counterpoint to 1 would be a boundary point. On the boundary, any circle you draw ( no matter how small ) will include some points from inside the domain and some outside.
Condition 1 indicates that the region is open. An example might be R2 with 0< x, y< 1. For any point near (0,0) or (1,1) there will exist a value r small enough that you can draw a circle around the point with only members of the region inside.
Thank you.
 
  • #4
HallsofIvy
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Yes. condition 1 is saying that the set does not contain any of its boundary points. The two sets [itex]\{ (x, y)| x^2+ y^2\le 1\}[/itex] and [itex]\{ (x, y)| x^2+ y^2< 1\}[/itex] have the same boundary: [itex]\{(x, y)| x^2+ y^2= 1\}[/itex], the circle with radius 1. The first contains that boundary and is a "closed" set. The second does not contain any point of that boundary and is called an "open" set.
 

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