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

  • #1

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
[tex]
U=\{x\in R^n : |x|\leq 1\}.
[/tex]
While it is evident that
[tex]
\{x\in R^n : |x|\lt 1\},
\{x\in R^n : |x|= 1\},
\{x\in R^n : |x|\gt 1\}
[/tex]
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 [itex] \{x\in R^n : |x|= 1\} [/itex] is the boundary of U; I need to show that if [itex] x [/itex] is any point in said set, and A is any open rectangle such that [itex] x\in A [/itex], then A contains a point in U and a point not in U. If x is such that [itex] |x|=1 [/itex], then [itex] x\in U [/itex], so I know that any open rectangle [itex] A [/itex] about the point[itex] x [/itex] contains at least one point in U (namely [itex]x[/itex]), how do I know my open rectangle [itex] A [/itex] also contains points for which [itex] |x|\gt 1 [/itex]?


Homework Equations



An open rectangle in [itex] R^n [/itex] is a set of the form [itex] (a_1,b_1)\times ... \times (a_n,b_n) [/itex].
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).

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
22,129
3,297
Hi SrEstroncio! :smile:

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.
 
  • #3
The points [itex]x\in R^n [/itex] for which any open rectangle [itex] A [/itex] with [itex] x\in A [/itex] contains points in both [itex] U [/itex] and [itex] R^n - U [/itex] are said to be the boundary of U.
 
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  • #4
22,129
3,297
The points [itex]x\in R^n [/itex] for which any open rectangle [itex] A [/itex] with [itex] x\in A [/itex] contains points in both [itex] U [/itex] and [itex] R^n - U [/itex] 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

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

such that

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

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

[tex]A^\prime\subseteq A[/tex]

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

So, try to work this out in detail. This should form the first step.
 
  • #5
Let [itex] R [/itex] be an open rectangle such that [itex] x \in R [/itex], [itex] R=(a_1,b_1)\times ... \times (a_n,b_n) [/itex]. If [itex] x=(x_1,...,x_n) [/itex], we construct an open rectangle [itex] R' [/itex] with sides smaller than [itex] 2\min{(b_i - x_i, x_i-a_i)} [/itex] for [itex]1\leq i \leq n [/itex], and centered about the point [itex] x [/itex]. By construction [itex] R' \subset R [/itex] and this construction can always be done.

To prove the set [itex] |x|=1 [/itex] is the boundary of U, I must take a point for which [itex] |x|=1 [/itex] and let R be any open rectangle containing [itex] x [/itex], I must now show that [itex] R [/itex] contains points both in [itex] U [/itex] and points which are not on [itex] U [/itex].
 
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  • #6
22,129
3,297
Let [itex] R [/itex] be an open rectangle such that [itex] x \in R [/itex], [itex] R=(a_1,b_1)\times ... \times (a_n,b_n) [/itex]. If [itex] x=(x_1,...,x_n) [/itex], we construct an open rectangle [itex] R' [/itex] with sides smaller than [itex] 2\min{(b_i - x_i, x_i-a_i)} [/itex] for [itex]1\leq i \leq n [/itex], and centered about the point [itex] x [/itex]. This construction can always be done.

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

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

such that

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

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

A slight problem however, the point [itex](a_1,...,a_n)[/itex] does not lie in our rectangle!! Can you solve this?
 
  • #7
Sorry for the inactivity, my computer decided to self-destruct under the heat.

Well, that [itex] (a_1,a_2,...,a_n) [/itex] does not lie in our rectangle centered about the point [itex] x [/itex] is not much of a problem, since said rectangle was constructed inside our original and arbitrary rectangle, not necessarily centered at [itex] x [/itex], and [itex] (a_1,a_2,...,a_n) [/itex] does lie in it.
 
  • #8
22,129
3,297
Sorry for the inactivity, my computer decided to self-destruct under the heat.

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

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

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