Determine if a set is open and/or closed and find its limit points, isolated points, and closure.

Question

Given the set $A=\{\frac{(-1)^n}{n}:n=1,2,3,...\}$, I need to determine if this set is open/closed, what its limit points are, what its isolated points are, and what is the closure of the set. My attempt at a solution is below.

I see that $A$ can also be written as $\{-1,\frac{1}{2},\frac{-1}{3},\frac{1}{4},\cdots\}$, or $\{-1\} \cup \{\frac{1}{2},\frac{-1}{3},\frac{1}{4},\cdots\}$. From this, I see that $-1$ is an isolated point because $-1 \notin \mathbb{Q}$ like the other elements of $A$ are. I also believe that $\frac{1}{2}$ is a limit point because if we take $V_{\epsilon}(\frac{1}{2})$, then we cannot go less than $\frac{1}{2}$ or else we will be outside $\mathbb{Q}$. With this in mind, the set must be open. It cannot be closed because the only sets that are both open and closed are $\mathbb{R}$ and $\emptyset$.

I'm a little unsure of the closure. I know that the definition of closure is $\bar{A}=A \cup L$ where $L$ is the limit point. So I suppose the closure of the set is $A \cup L=\{\frac{(-1)^n}{n}:n=1,2,3,...\} \cup \{\frac{1}{2}\}$. But I'm confused because $\frac{1}{2} \in A$.

Answer 1

This set (as a subset of $\mathbb{R}$) is neither open, nor closed. The closure of $A$ is $A\cup \{0\}$ since 0 is the only limit point of the set. All points are isolated.

$A$ is not open: $B_{r}(x)\not\subset A$ for any $r>0$ and any $x\in A$.

$A$ is not closed: $0$ is a limit point of $A$ but is not contained in $A$ (I will show it is a limit point next).

Limit points: A limit point of a set is a point $x$ such that there exists $x_n\in A$ such that $x_n\to x$. I leave it as an exercise for you to show that 0 is the only point satisfying this.

Isolated points: Around any point $x\in A$, the neighborhood $B_{1/n^3}(x)\cap A=x$ (here $x=(-1)^n/n$)

Answer 2

Okay, big picture: You have a set $A$ and you have a point $p$ that may, or may not, be in $A$.

We consider neighborhoods around $p$ and if we take small enough neighborhoods one of three things can happen:

1) it could be that if we take a neighborhood small enough, that the neighborhood will not contain any points of $A$ (other than possibly $p$ itself). If $p$ is actually in $A$ then $p$ is an isolated point. (It is isolated from any other point is $A$). (If $p$ isn't in $A$ then this is just a point that has nothing to do with $A$; there isn't any necessary term for it.)

2) it could be that every neighborhood we take, no matter how small, will always contain a point point of $A$ (other than $p$ itself). This is the exact opposite of 1) and 1) and 2) are utterly contradictory and mutually exclusive. This is called a limit point. (The point sits at a limit of $A$.) Now if $p$ is a limit point it may or may not be that $p$ is a point of $A$.

3) it could be that there is a neighborhood small enough so that every point in the neighborhood, including $p$ itself, is in $A$. This is called an interior point. (Such a point is nestled deeply in the interior of $A$.) Now 3) contradicts 1) but they aren't mutually exclusive; it could be possible that a point is neither 1) nor 3). And 3) does not contradict 2). In fact, if 3) is true then $p$ is also a limit point as well as an interior point. All interior points are limit points but limit points don't need to be interior points.

$\mathbb Q$ has nothing to do with this question whatsoever. The only set we are concerned with is $A$.

So what of $\frac 12$. Which of these three cases apply.

1) if you take an neighborhood small enough, say $\epsilon = \frac 14$ then $V_{\frac 14} (\frac 12) =(\frac 14, \frac 34)$ does not have any points in $A$ other than $\frac 12$ itself. So $\frac 12$ is not a limit point.

2) $V_{\frac 14}(\frac 12) = (\frac 14, \frac 34)$ is a neighbor hood that contains no point of $A$ other than $\frac 12$ so 2) holds. And as $\frac 12 \in A$, $\frac 12$ is an isolated point.

3) If we take any neighborhood with a radius smaller than $\frac 14$ around $\frac 12$, that neighborhood will always have points not in $A$. So There is no neighborhood around $\frac 12$ that is completely in $A$. So $\frac 12$ is not an interior point.

So are there any limit points?

Is there some $p\in \mathbb R$ ($p$ need not be in $A$) so that every neighborhood contains a point of $A$. Is there some point that exists on the "limit" of $A$?

.... that's enough for now...

consider what this all means and then reread the definition of closed. (Every limit point is in $A$. Is the limit point I hinted at in $A$. If not.... then $A$ is not closed.)

Reread the definition of open. (Every point in $A$ is an interior point. Well, $\frac 12 \in A$ ... and $\frac 12$ is not an interior point.... so can $A$ be open. [Hint: Of course not!])

Reread the definition of closure. (The set that contains all the point of $A$ and also contains all the limit points. So $A \cup \{$ the limit points of $A$ that I hinted at$\}$.