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
$$U=\{x\in R^n : |x|\leq 1\}.$$
While it is evident that
$$\{x\in R^n : |x|\lt 1\},<br /> \{x\in R^n : |x|= 1\},<br /> \{x\in R^n : |x|\gt 1\}$$
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 point$x$ 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 $R'$ 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 $R' \subset R$ 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 $R'$ 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...