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

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 $$\frac{1}{2} \in A$$.

• There are many mistakes in your argument so far. First of all, $-1$ is in $\mathbb{Q}$. But that has no bearing on whether it's an isolated point of $A$. Commented Apr 1, 2020 at 18:12
• Um, of course negative one is a rational number! And even if it wasn't what does that have to do with whether it is an isolated point or not. "We can not go less than $\frac 12$ or we are outside $\mathbb Q$". A limit point means no matter how small you go you will find at least one point in the set $A$ so that is not a good argument. And why are you talking about $\mathbb Q$? $\mathbb Q$ has nothing to do with anything. .... I really hate to say this but there is absolutely nothing correct or salvageable in this. I really, really advise seeing your professor in office hours. Commented Apr 1, 2020 at 18:25

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 $$x=(-1)^n/n$$)

• Thanks for your answer. Can you explain how you arrived at each conclusion? I was taught these concepts yesterday and I want to learn as much as possible. Commented Apr 1, 2020 at 18:23

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$$\}$$.