Finding border points and then interior points

[A.Magnus]

I am working on a classical real analysis problem as follow:

Find $int (A)$ and $bd (A)$ if $A = \{ \frac{1}{n} | n \in \mathbb N \}= \{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ... \}$.

The answers from solution manual are respectively $ int (A) = \emptyset$ and $bd (A) = \{0\} \cup \{ \frac{1}{n} | n \in \mathbb N \}$. And here are my textbook's definition of interior point and border point:

Let $A \subseteq \mathbb R$. A point $x$ in $\mathbb R$ is an interior point of A if there exists a neighborhood $N$ of $x$ such that $N \subseteq A$. If for every neighborhood of $N$ of $x$, $N \cap A \neq \emptyset$ and $N \cap A^c \neq \emptyset$, then $x$ is a border point of $A$.

And then there is also this:

Every point $x \in A$ is either interior or border point of $A$.

To my inexperienced mind, the answer to $int (A) = \emptyset $ is understandable because while neighborhood $N (x | \epsilon >0)$ is an open set (per textbook), but $A = \{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ... \}$ is incomplete, meaning that there is gap between any two consecutive elements of $A$. Therefore by technicality $N (x | \epsilon )$ does not exist inside $A$. (Am I correct here?)

The answer to $bd (A) = \{0\} \cup \{ \frac{1}{n} | n \in \mathbb N \}$ is understandably the direct consequence of $int (A) = \emptyset $, i.e., if none of the elements of $A$ is interior point, then all of them must be border points, plus the zero.

But what troubles me is this: What happens if I instead answer $int(A)$ first and then $bd (A)$ second? Am I going to get the same answers?

(i) For $\{0\}$, I was tempted to conclude that zero is indeed a border point because $N (0 | \epsilon) \cap A \neq \emptyset$ and $N (0 | \epsilon) \cap A^c \neq \emptyset$ for all $\epsilon >0$. But had I not just concluded above that $N (x | \epsilon)$ does not exist in $A$?

(ii) How about for the rest of the points, i.e., $\frac{1}{n}$? How do I conclude $N (\frac{1}{n} | \epsilon ) \cap A \neq \emptyset$ and $N (\frac{1}{n} | \epsilon ) \cap A^c \neq \emptyset$ for all possible $\epsilon >0$?

I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time and gracious help. ~MA

 

[Olinguito]

Hi MaryAnn.

Therefore by technicality $N (x | \epsilon )$ does not exist inside $A$. (Am I correct here?)

You mean to say that for all $\epsilon>0$, $N(x\mid\epsilon)\not\subseteq A$, because $N(x\mid\epsilon)$ contains points not in $A$. For example, for $x=\frac13$, given $\epsilon>0$, let
$$\delta=\min\left\{\frac{\epsilon}2,\frac{\frac13-\frac14}2\right\};$$
then $y=x-\delta\not\in A$ but $y\in N(x\mid\epsilon)$. (If all this looks confusing to you, try sketching the points on a number line.) In general, if $x=\frac1n$, let
$$\delta=\min\left\{\frac{\epsilon}2,\frac{\frac1n-\frac1{n+1}}2\right\};$$
and $y=x-\delta$ as above. Hence there does not exist any $x$ such that $x$ is an interior point of $A$ (not that $N (x | \epsilon )$ does not exist).

But what troubles me is this: What happens if I instead answer $int(A)$ first and then $bd (A)$ second? Am I going to get the same answers?

You can do whichever one first, it doesn’t really matter. If you do everything correctly, you should get the same answers.

(i) For $\{0\}$, I was tempted to conclude that zero is indeed a border point because $N (0 | \epsilon) \cap A \neq \emptyset$ and $N (0 | \epsilon) \cap A^c \neq \emptyset$ for all $\epsilon >0$. But had I not just concluded above that $N (x | \epsilon)$ does not exist in $A$?

See above. Note also that in the above you were taking $x\in A$ whereas now you are considering $0$, which is not in $A$.

(ii) How about for the rest of the points, i.e., $\frac{1}{n}$? How do I conclude $N (\frac{1}{n} | \epsilon ) \cap A \neq \emptyset$ and $N (\frac{1}{n} | \epsilon ) \cap A^c \neq \emptyset$ for all possible $\epsilon >0$?

For $x\in A$, $N (x | \epsilon)\cap A$ is clearly not going to be empty because it contains $x$ itself. It also contains points not in $A$ by the method outlined in (i) above. Hence every point in $A$ is a border point. For points outside $A$, $0$ is the only point that is a border point. Given any $\epsilon>0$, $N(0\mid\epsilon)$ contains a point in $A$ (namely $\frac1N$ where $N$ is any integer greater than $\frac1\epsilon$) and also a point not in $A$ (namely $0$ itself). For any $x\not\in A$ and $x\ne0$, let
$$\epsilon=\tfrac12\inf\left\{\{|x|\}\cup\{|x-a|:a\in A\}\right\};$$
then $N (x | \epsilon)\cap A$ does not contain any point in $A$ and so $x$ can’t be a border point.