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Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Set problem

Last edited by Stanley_Marsh (2007-04-09 09:18:56)

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Set problem

The proof is like this :
 

Here is what I dont understand, When he defined

That h is positive , then q is already an interior point , why does he need to make it so complicated.Or I miss something?

Last edited by Stanley_Marsh (2007-04-09 09:29:59)

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Numbers are the essence of the Universe

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Set problem

1. Just take an open ball in R² centered at some point x ∈ (a, b). No matter what the radius of the ball, it will always contain points not in (a, b). Therefore there exists no 2-ball contained entirely in (a, b), so (a, b) is not open in R². This is because the open ball in R² is (in a standard sense) a circle, while the open ball in R is just an open interval.

2. He needed to "make it more complicated" because he cannot just say that since h is positive, q is an interior point. Imagine h getting smaller and smaller so that q is closer and closer to being a boundary point, so that d(p, q) approaches r. He does the other steps to show that q is never a boundary point. How can you reason this? Well he shows that any point s in an open ball of radius h centered at q is contained in E. Since q is arbitrary, we see that for any point in E this holds. This is precisely the definition of openness; that at any point of the set E you can center an open ball B of some suitable radius such that the ball is entirely contained in the set; so B is a subset of E. The proof just shows that the suitable radius is this h.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Set problem

1. Just take an open ball in R² centered at some point x ∈ (a, b). No matter what the radius of the ball, it will always contain points not in (a, b). Therefore there exists no 2-ball contained entirely in (a, b), so (a, b) is not open in R². This is because the open ball in R² is (in a standard sense) a circle, while the open ball in R is just an open interval.

Oddly enough, the phrase "open ball" only really makes sense in R³

But that reminds me of a joke.  A mathematician and an engineer walk out of a conference, and the engineer remarks, "I find it so difficult picturing all this stuff in 11 dimensional space." to which the mathematician replies, "I just picture it in n-dimensional space, then set n equal to 11."

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Set problem

I guess it's just another one of those "contradictory mathematical terminologies." Even if we revised it to be called a "open (n-1)-sphere" it isn't necessarily (n-1)-spherical region in a general n-dimensional metric space, as the metric can certainly give it a different shape. Oh well, the term has stuck with me.

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Set problem

O, but that's very hard to come up with such prove.

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Numbers are the essence of the Universe