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I want to prove {$x\in \textbf{R}^n| |x-a|=r$} $\Rightarrow x\in \partial B_r(a)$. And I know how to prove the inverse by using triangle inequality.

Where $B_r(a)$={$z\in R^n| |z-a|

$u$ is boundary point of A iff: $\forall \epsilon>0,$ $B_\epsilon(a)$ $\cap A \neq \phi $ and $B_\epsilon(a)$ $\cap (R^n$\ $A$)$\neq \phi $

and $\partial A$ is the set of all boundary point of A.

But to prove this way, it seems that I cannot apply the triangle inequality. How can I use the definition that any open ball centered by a boundary point of A intersects both int and ext of A?

  • 1
    Please state your definitions of $B_r(a)$ and $\partial B_r(a)$. – 2017-02-23
  • 0
    Possible duplicate of [How to show $\partial B_r(a)$={$x\in R^n| |x-a|=r$}](http://math.stackexchange.com/questions/2150851/how-to-show-partial-b-ra-x-in-rn-x-a-r) – 2017-02-23

1 Answers 1

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Suppose $|x-a|=r$.

To prove that $x \in \partial B_r(a)$ you must prove that for every $\epsilon>0$ the sets $B_\epsilon(x) \cap B_r(a)$ and $B_\epsilon(x) \cap (\mathbb{R}^n - B_r(a))$ are non-empty.

I'm going to prove nonemptiness just of the set $B_\epsilon(x) \cap B_r(a)$; once you get the idea, I suspect you'll get the other one.

By definition, $$B_\epsilon(x) \cap B_r(a) \ne \emptyset \,\text{if and only if}\, \exists q \in B_\epsilon(x) \cap B_r(a) $$ As with any existence proof, to prove it you first use your intuition, or knowledge, or a picture, or anything you can use to help you, in order to make an educated guess at what the correct and appropriate point $q$ could possibly be. Using that intuition, you must then define $q$ precisely. Finally, using your definition of $q$, you must prove that $q \in B_\epsilon(x) \cap B_r(a)$.

Here's the intuition that tells me what to do: if I move straight inward from $x$ along a radius of $B_r(a)$, I will be in $B_r(a)$; and if move inward just a tiny bit, I will also be in $B_\epsilon(x)$.

Now let's make the intuition precise in order to arrive at a precise construction of the point $q$. Moving inward from $x$ along a radius of $B_r(a)$ is the same as moving along the straight line segment $$(1-t)x + ta, \quad 0 \le t \le 1 $$ If $t$ is small enough, the distance from $q=(1-t)x+ta$ should be less than $\epsilon$. In fact, I can probably arrange that the distance is an exact positive value less than $\epsilon$ such as $\epsilon/2$. What value of $t$ would make the distance to $x$ equal to $\epsilon/2$? That's a concrete question I can answer: $$|q-x| = \epsilon/2 $$ $$|((1-t)x + ta) - x| = \epsilon/2 $$ $$|-tx+ta| = \epsilon/2 $$ $$t \cdot |a-x| = \epsilon / 2 $$ $$t r = \epsilon/2 $$ $$t = \frac{\epsilon}{2r} $$ So I set $t$ equal to that value, define $t = (1-t)x + ta$, and work my prove backwards to prove that $|q-x| = \epsilon/2$ and therefore $q \in B_\epsilon(x)$. And I can also use an exact calculation to prove that $|q-a| < r$.