Understanding the boundary of a set

Question

I try to understand the boundary of a set.

I know the definition (Let $A \subseteq \mathbb{R} $: P is a point of the boundary, if for every small $\epsilon \in \mathbb{R}, \epsilon>0$ are points in $A$ and in $A^c$.

I understood, why the boundary of $(0,1]$ is $\lbrace 0,1 \rbrace$.

Now I'm struggling with $M=\lbrace \frac{1}{n}, n \in \mathbb{N} \rbrace$. I know the solution $\partial M =M \cup \lbrace 0 \rbrace$.

But I don't understand the solution. If I choose for example $p=\frac{1}{2}$ and $\epsilon=\frac{1}{100}$, then I don't see, why $\frac{51}{100}$ is in $A$ and $\frac{-49}{100}$ is in $A^c$

Perhaps someone can help me out?

Answer 1

I'm not entirely sure about your current definition of boundary, but it seems like you want to say that $a\in\mathbb{R}$ is in the boundary of $A\subseteq\mathbb{R}$ iff for all $\epsilon>0$, both $(a-\epsilon,a+\epsilon)\cap A$ and $(a-\epsilon,a+\epsilon)\cap A^c$ are non-empty.

Just consider a point $a\in\mathbb{R}$ and $A=\{a\}$. Then $a$ will always be an element of $(a-\epsilon,a+\epsilon)$, independent of $\epsilon$, so this interesection will be non-empty, while the intersection with $A^c$ won't be empty either, since it is $(a-\epsilon,a)\cup(a,a+\epsilon)$.

Now consider what happens if you add a point, say $b$ to $A$. If we have $\epsilon<|b-a|$, there's no problem at all and you can do the same. For $\epsilon>|b-a|$, you still find that $a\in (a-\epsilon,a+\epsilon)\cap A$ and any point inbetween $a$ and $b$ will be in both the interval and $A^c$ as well.

In a similar way, it is possible to show in your example that $M\subseteq \partial M$. You still need to show that $0\in\partial M$. Now we have an arbitrary $\epsilon$ given and this seems to be where your confusion arises. $\epsilon$ is arbitrary, yet fixed. We do not know what it is, just that is there and won't change from this point onward. We then know that there is a natural number $n\in\mathbb{N}$ such that $\frac{1}{n}<\epsilon$ and thus $\frac{1}{n}\in(-\epsilon,\epsilon)$. The intersection of this interval with $M$ thus isn't empty. Moreover, all negative elements in the interval are obviously in $M^c$ and so the intersection with $M^c$ isn't closed either, so $0\in\partial M$ as well.

I hope this clarifies things a bit. If not, please ask!

Answer 2

If $p=\frac 1n$ then $p\in\partial M$ because: No matter, how small you choose $\epsilon>0$, the point $p$ itself is in $B_\epsilon(p)\cap M$. And the point $p-a$ where $0<a<\min\{\epsilon, \frac1{p(p+1)}\}$ is in $B_\epsilon(p)\cap M^c$ (why?)

If $p=0$, then for any $\epsilon>0$ we find $0\in B_\epsilon(0)\cap M^c$ and $\frac1n\in B_\epsilon(0)\cap M$ for any $n>\frac1\epsilon$.

For any other point $p\in\mathbb R$ we can show that $p\notin\partial M$ by exhibiting an $\epsilon>0$ such that $B_\epsilon(p)\cap M=\emptyset$:

• If $p<0$, let $\epsilon=-p$.
• If $p>1$, let $\epsilon=p-1$.
• If $0<p<1$, but there is no $n$ with $p=\frac1n$, let $n=\lfloor \frac1p\rfloor$ (note that $\frac1p>1$). Then $\frac1{n+1}<p<\frac1n$ and we can take $\epsilon=\min\{\frac1n-p,p-\frac1{n+1}\}$.