R^2 open and closed?

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If we define a set, c, to be R^2, how is it open and closed?

The definitions I'm using:

Open set: Open set, O, is an open set if for all points x are in O, and we can find ONE B(x,ρ) such that B(x,ρ) is less than zero.

Closed set: Compliment of an open set, AKA R^n/O.

This isn't a HW question, I'm reviewing the analysis part of my PDE course from fall semester, and this is something extra she told us, and now I don't know how this could be true!

Thanks for the help.
 
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  • #2
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"Openness" and "closedness" aren't mutually exclusive, despite their unfortunate names. This is obvious topologically (the whole space is open by definition, but it is also the complement of the (open) empty set, and so it is also closed), but there's no need to abstract as far as topology with Rn; that every point in R2 is an interior point (has an open ball in R2) in should be obvious, so it is open. But R2 also contains all of its limit points (why?), so it is closed.
 
  • #3
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But R2 also contains all of its limit points (why?), so it is closed.
Ok, well that's probably why she said it was beyond the scope of the PDE course. Guess I'll do some reading on limit points!

Thanks for your help!
 
  • #4
HallsofIvy
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If we define a set, c, to be R^2, how is it open and closed?

The definitions I'm using:

Open set: Open set, O, is an open set if for all points x are in O, and we can find ONE B(x,ρ) such that B(x,ρ) is less than zero.
This looks like non-sense to me. What does it mean for a set to be "less than zero"? I think you mean "we can find at least one B(x,ρ) that is a subset of O". That's obviously true for R2 since, for any x in R2 B(x,ρ) is certainly a subset of R2.

Closed set: Compliment of an open set, AKA R^n/O.
R2 is the compliment of the empty set so it is sufficient to prove that the empty set is open. And that follows from the logical principal that "if P then Q" is true in the case that P is false, no matter whether Q is true of false. For the empty set, "if x is in O" is always false because the empty set contains NO points.

This isn't a HW question, I'm reviewing the analysis part of my PDE course from fall semester, and this is something extra she told us, and now I don't know how this could be true!

Thanks for the help.[/QUOTE]
 
  • #5
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This looks like non-sense to me. What does it mean for a set to be "less than zero"? I think you mean "we can find at least one B(x,ρ) that is a subset of O". That's obviously true for R2 since, for any x in R2 B(x,ρ) is certainly a subset of R2.
Hmm, this is out of my notes, so maybe I put zero, when I really meant O as I never understood why the ball at some element would have to be less than 0. My friend is borrowing my book right now, but I'd be able to check what the book says once I get it back.

x is an element, and ρ is the radius.

The book is Introductory to Patrial Differential Equations with Applications by Zachmanoglou and Thoe, if anyone has the book and can check.
 

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