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I started studying multivariable caclulus, and am having problems with this exercise:

Given an infinite bounded set A in $R^n$,$2\leq n$, show there are infinite boundary points

Attempt: Going from the assumption that A has finite boundary points, we can assume it has an interior point. I believe it's possible to construct a ball around that point with large enough radius (assuming A is bounded by M, 3M should suffice), and prove that there's a boundary point on every line between a boundary point of that ball and the interior point we selected. This ended up being very complex and I didn't manage to work out all the details nicely, so I figure there must be something easier!

Thanks!

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2 Answers 2

For the $n=2$ case:

Let $A$ be such a set and let $L_x$ denote the vertical line that passes through $(x,0)$. Note that any boundary point of $L_x\cap A$ (as a subset of $L_x$) is a boundary point of $A$, as if an open ball $B$ contains a point in $L_x$ outside $A\cap L_x$ then this point lies outside $A$ as well. Since $A$ is bounded, for each $x\in\mathbb R$ either $L_x\cap A$ is empty or it has at least one boundary point (this is not hard to show). If there are infinitely many $x\in \mathbb R$ such that $L_x\cap A$ is nonempty, then we have infinitely many distinct boundary points and are done. Otherwise, for each $x$ such that $L_x\cap A$ is nonempty we have some $\epsilon>0$ such that $L_y\cap A$ is empty for any $y$ such that $|x-y|<\epsilon$. Thus any point $p\in L_x\cap A$ is such that $B_r(p)$ contains $B_{\min\{r,\epsilon\}}(p)\setminus L_x$ which is nonempty and does not intersect $A$, so is a boundary point of $A$. Hence all points in $A$ are boundary points, so $A$ has infinitely many boundary points.

For $n>2$, the argument is very similar. I will let you work out the details.

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I think the part you said is not hard to show is exactly where I'm stuck in. How might we show that the intersection must contain a boundary point? Intuitively it's obvious, but... I can't see how to prove it. Thanks! –  ro44 Mar 23 '12 at 1:56
    
Well, the intersection is a subset of the line $L_x$, so we can treat it as $\mathbb R$. It thus has a least upper bound and a greatest lower bound, both of which can be shown to be limit points. –  Alex Becker Mar 23 '12 at 2:02
    
Thanks, I got it now. Now I guess I'm not entirely sure how to generalise $L_r$. It looks like it should be a plane if $n=3$? Is this the correct direction? –  ro44 Mar 23 '12 at 2:22
    
You could make an inductive argument like that, but I suspect it'll be messy. How about you consider lines $L_{x_1,x_2,\ldots,x_{n-1}}$ which have all but one coordinate fixed for the case $n>2$? –  Alex Becker Mar 23 '12 at 2:26
    
I'll try that, thanks. –  ro44 Mar 23 '12 at 2:39
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If $A$ has no interior point then every point of $A$ is on the boundary so there are infinitely many boundary points. Otherwise $A$ contains two points such that the line $L$ which connects them is contained in $A$. Let $v$ be a nonzero vector that is orthogonal to $L$ (this exists since $n \geq 2$). For every point $x \in L$ consider the set $ \{\lambda \geq 0: x+\lambda v \in A\}$. It is non-empty because it contains $0$ and it is bounded because $A$ is bounded. Let $\lambda_0$ the supremum of this set. Then $x+\lambda_0 v$ is a boundary point of $A$. The lines $x + \mathbb Rv$ ($x \in L$) are disjoint so $A$ has infinitely many boundary points.

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I've considered something similar. My problem is actually showing that the point defined by the supremum of the set is a boudnary point (we defined boundary points as points such that every ball centered around them always contains both interior and exterior points). How might this be proven? Thanks! –  ro44 Mar 23 '12 at 1:56
    
If $B$ is a ball centered around $x+\lambda_0 v$ it contains points $x+(\lambda_0+\varepsilon)v$ and $x+(\lambda_0-\varepsilon)v$ for some small $\varepsilon > 0$. Since $\lambda_0$ is an upper bound for the set of $\lambda$s with $x+\lambda v \in A$ the point $x+(\lambda_0+\varepsilon)v$ is not in $A$. And since $\lambda_0$ is the least upper bound $\lambda_0-\varepsilon$ can not also be an upper bound so there is some $\lambda_0-\varepsilon < \mu \leq \lambda_0$ such that $x+\mu v$ is in $A$. This shows that $x+\lambda_0 v$ is indeed a boundary point. –  marlu Mar 23 '12 at 11:54
    
Why is there a nonzero vector that is orthogonal to L, and why do you need it to be orthogonal? –  daniel.jackson Apr 4 '12 at 21:29
    
It doesn't need to be orthogonal, it is sufficient that $v$ and the directional vector $w$ of $L$ are linearly independent, so you can conclude that the lines $x+\mathbb R v$ ($x\in L$) are disjoint: If $x_0$ is any point of $L$ and $x_0+a w+\lambda v = x_0+b w+\mu v$ ($a,b,\lambda,\mu \in \mathbb R$) then $a=b$ and $\lambda=\mu$ by linear independence. The existence of $v$ follows from $n\geq 2$: the vector $w$ spans a subspace of dimension 1 so it is not the whole space and any vector $v$ not in $\mathbb R w$ will be linearly independent of $v$. The intuition is easy if you draw a picture. –  marlu Apr 13 '12 at 20:07
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