Boundary of closed sets (Spivak's C. on M.)

[SrEstroncio]

Homework Statement

I have been self studying Spivak's Calculus on Manifolds, and in chapter 1, section 2 (Subsets of Euclidean Space) there's a problem in which you have to find the interior, exterior and boundary points of the set
<br /> U=\{x\in R^n : |x|\leq 1\}.<br />
While it is evident that
<br /> \{x\in R^n : |x|\lt 1\},<br /> \{x\in R^n : |x|= 1\},<br /> \{x\in R^n : |x|\gt 1\}<br />
are the interior, boundary and exterior of U, in that order, I am stuck proving it. In particular, I can't quite grasp how to prove rigorously that the set \{x\in R^n : |x|= 1\} is the boundary of U; I need to show that if x is any point in said set, and A is any open rectangle such that x\in A, then A contains a point in U and a point not in U. If x is such that |x|=1, then x\in U, so I know that any open rectangle A about the pointx contains at least one point in U (namely x), how do I know my open rectangle A also contains points for which |x|\gt 1?

Homework Equations

An open rectangle in R^n is a set of the form (a_1,b_1)\times ... \times (a_n,b_n).
Spivak defines interior, exterior and boundary sets using open rectangles, not open balls.

The Attempt at a Solution

It is obvious that the boundary of the n-ball is the n-sphere, and most books wouldn't bother proving it, but I like to be rigorous in my proofs. I am getting stuck in the technical details (how do I know not all points in my open rectangle are equidistant from the origin?, how do I know at least one is "farther away?", that kinda stuff).

 

[micromass]

Hi SrEstroncio!

It's good that you want to be rigorous about such a things. So let's see if I can help you prove this.

First, I would like to hear from you how Spivak defined boundary in terms of open rectangles.

 

[SrEstroncio]

The points x\in R^n for which any open rectangle A with x\in A contains points in both U and R^n - U are said to be the boundary of U.

 

[micromass]

The points x\in R^n for which any open rectangle A with x\in A contains points in both U and R^n - U are said to be the boundary of U.

OK, that definition is slightly uglier than I had hoped for. So we will costumize it a bit. Can you prove that we can take x the center of the rectangle?

That is, we can take

A=[a_1,b_1]\times...\times [a_n,b_n]

such that

x=(\frac{b_1-a_1}{2},...,\frac{b_n-a_n}{2})

How should we prove such a thing? Well, we might might find a rectangle

A^\prime\subseteq A

such that A' has the property that x is the center of the rectangle. Now, if we can prove that A' interesects U and \mathbb{R}^n\setminus U, then A also intersects these sets.

So, try to work this out in detail. This should form the first step.

 

[SrEstroncio]

Let R be an open rectangle such that x \in R, R=(a_1,b_1)\times ... \times (a_n,b_n). If x=(x_1,...,x_n), we construct an open rectangle R&#039; with sides smaller than 2\min{(b_i - x_i, x_i-a_i)} for 1\leq i \leq n, and centered about the point x. By construction R&#039; \subset R and this construction can always be done.

To prove the set |x|=1 is the boundary of U, I must take a point for which |x|=1 and let R be any open rectangle containing x, I must now show that R contains points both in U and points which are not on U.

 

[micromass]

Let R be an open rectangle such that x \in R, R=(a_1,b_1)\times ... \times (a_n,b_n). If x=(x_1,...,x_n), we construct an open rectangle R&#039; with sides smaller than 2\min{(b_i - x_i, x_i-a_i)} for 1\leq i \leq n, and centered about the point x. This construction can always be done.

Indeed, that's a nice first step. So our situation now is that we have a rectangle

]a_1,b_1[\times...\times ]a_n,b_n[

such that

x=(\frac{b_1-a_1}{2},...,\frac{b_n-a_n}{2})

Now we want to find a point in U. Well, an obvious choice is (a_1,...,a_n). Can you prove that this is in U?

A slight problem however, the point (a_1,...,a_n) does not lie in our rectangle! Can you solve this?

 

[SrEstroncio]

Sorry for the inactivity, my computer decided to self-destruct under the heat.

Well, that (a_1,a_2,...,a_n) does not lie in our rectangle centered about the point x is not much of a problem, since said rectangle was constructed inside our original and arbitrary rectangle, not necessarily centered at x, and (a_1,a_2,...,a_n) does lie in it.

 

[micromass]

Sorry for the inactivity, my computer decided to self-destruct under the heat.

Well, that (a_1,a_2,...,a_n) does not lie in our rectangle centered about the point x is not much of a problem, since said rectangle was constructed inside our original and arbitrary rectangle, not necessarily centered at x, and (a_1,a_2,...,a_n) does lie in it.

Indeed, so that wouldn't pose a problem...