# Closed Sets and Open Sets

I have a few questions regarding open and closed sets. I am given a set: $$A = \left\{ \frac{1}{x}: x \in \mathbb{Z}^+ \right\},$$ I was asked to find the interior, closure, and boundary points.

This is my attempt:

Interior: $( 0,+ \infty)$

Boundary: $\{0\}$

Closure: $[0, +\infty)$

I have a feeling I am doing this completely wrong..

While I was looking up help, I noticed that a lot of people has been asking for the boundary, closure and interior points of $\sin(1/x)$, but I cannot why they all said there are no interior, and all points are in the boundary.

Thanks!

• The interior of $A$ is the union of all open subsets contained in $A$ and hence a subset of $A$,but $(0,\infty)\not\subseteq A$.
– user149890
May 22, 2014 at 18:21
• is $Z+$ the set of positive integers? If it is, then you just have a bunch of isolated points in $\mathbb{R}$, no interior. But you do have one limit point... May 22, 2014 at 18:25
• Thanks for help! Can you give me an example of a different function with interior points? Sorry I haven't fully grasp this material yet. When my prof explained it with sets such as (1,5) it made sense but I don't quite understand what to do with functions! May 22, 2014 at 18:33
• There is no such thing as interior/closure/boundary of a function, just for sets. You mean probabla the following: For every function $f:X\rightarrow Y$ you could consider the graph $\lbrace (x,y)\in X\times Y: y=f(x) \rbrace\subset X\times Y$. This is a set and therefore it's possible to speak of its interior/closure/boundary.
– user149890
May 22, 2014 at 18:37

First, we can see that the interior of $A$ is the empty set, as $A$ consists of isolated points. In particular, for any open interval $(a,b)\subset \mathbb{R}$, $(a,b)\not\subset A$ as $(a,b)$ contains irrational numbers. Note that $\lim\limits_{n\to\infty}\frac{1}{n}=0$, and so $0\in \overline{A}$. $A$ can be seen as the set of elements in the Cauchy sequence $\{1,\frac{1}{2},\frac{1}{3},\cdots\}$. Since limits of Cauchy sequences are unique, we know that $0$ is the only element in $\overline{A}$ that is not contained in $A$. Thus, $\overline{A}=A\sqcup \{0\}$. Finally, the boundary of $A$ is defined to be the set of points in the closure of $A$ not in the interior of $A$. Since the interior of $A$ is empty, we have that $\partial A=\overline{A}\setminus \emptyset=\overline{A}$.

To sum up: $A^o=\emptyset$ and $\partial A=\overline{A}=A\sqcup\{0\}$.

• Thank you! I think I understand now, I completely forgot that i would be including irrationals and rationals. If I change x in Z to x in R, since R is dense, would the interior points now be A? the boundary be 0, and the closure be A with 0? May 22, 2014 at 18:57
• Yes. To be more precise, you could change the definition of $A$ to be $A=\{\frac{1}{x}\,\mid\,x\neq 0\}$. Note that this set can be written as $A=(-\infty,0)\sqcup(0,\infty)$. Since $A$ is open, it is equal to its interior. It is clear that the boundary is $\{0\}$ and the closure is $\mathbb{R}$. May 22, 2014 at 18:59
• Thank you! I appreciate your help! I think I understand now! May 22, 2014 at 19:03

Take the set [0,1) as an example.

Interior = (0,1)

The interior of a set is the union of all open subsets of that set. Open subsets of [0,1) will look like: (a,b) where 0

Boundary = {0,1}

The boundary of a set is just all the boundary points. A boundary point is usually defined as a point where every neighborhood around the point contains points that are in the set AND points that are not in the set. In our example, 0 and 1 are the only two boundary points. Every neighborhood of 0 and 1 will intersect [0,1) but will also have points not in [0,1).

Closure = [0,1]

The closure is the original set plus all of it's boundary points. So when you add in the boundary points 0 and 1 to the original set of [0,1), you get [0,1].

{$\frac 11$, $\frac 12$, $\frac 13$, ...}