Is the following subset of $\mathbb{R}^2$ open, closed or neither?

Question

I am given the subset $A=\{(x, \sin(1/x)) | x>0\} \bigcup \{(0,y) | y \in [-1,1] \}\subset \mathbb{R}^2$ equipped with the standard Euclidean metric

I came to the conclusion that it is not open because for $x>$[some finite value>0], we clearly just have a 1 dimensional line in $\mathbb{R}^2$, about which none of the points have an open ball of finite radius that is entirely contained within the subspace (line).

To determine if it is closed, I have to consider whether its complement is open. On the one hand it is true that any point I can state in $\mathbb{R}^2$ (i.e. write the numbers down for) that is not contained in $A$, I can find an open ball around that is not contained in $A$, even if it is very close to $x=0$. I also cannot think of how I would construct a sequence in$A$ whose limit point is not in $A$, and of course if $L(A)=A$, then $A$ is closed.

However I am having trouble reconciling this with the fact that the sinusoidal squiggles become infintitely close as $x\rightarrow 0$. It is easy to show that the distance between adjacent intersections of the $y=0$ axis $\rightarrow 0$ as $x\rightarrow 0$. It makes it feel like the complement shouldn't really be open...

Would I be correct in thinking that the fact that the definition hinges on stating a specific element in the set or complement is the significant point here?

Also, I wanted to verify that $A=\{(x,y)| x\in \mathbb{Q}, y^n = x $ such that $ n\in \mathbb{Z}^+\}$ is neither open nor closed.

Answer 1

You are correct that $A$ is not open.

The set $A$ is also closed. As far as I can tell, there is no straightforward argument which follows directly from the definitions. Here is a possible alternative:

Recall that if $f(x,y)$ is continuous, then the set $f(x,y)=0$ is closed. Now consider the function $f(x,y)=x\sin(1/x)-xy$. It's fairly straightforward to show that $x\sin(1/x)$ is continuous, despite the squiggle. Then to solve $f(x,y)=0$, note that $0=(\sin(1/x)-y)x$, so either $x$=0 or $\sin(1/x)-y=0$. Thus the zero set of $f(x)$ is $A\cup\{(x,0)|y\in\mathbb{R}\}$. Now $A$ is the intersection of this set with the closed set $\{(x,y)|x\in\mathbb{R},y\in[-1,1]\}$. Recall that the intersection of two closed sets is closed. Thus $A$ is closed.

Since your second question is unrelated to the first, I would recommend deleting it and reposting it as a second question.

Answer 2

The set $A$ is closed in $\mathbb{R}^2$. Perhaps the easiest way to see this is to consider any $(x',y') \in \overline{A}$. We shall show that $(x',y') \in A$.

Since $\mathbb{R}^2$ is a metric space, there exists a sequence $(x_n,y_n) \in A$ such that $(x_n,y_n) \to (x',y')$, i.e. $x_n \to x'$ and $y_n \to y'$. Since all $x_n \ge 0$, we obtain $x' \ge 0$. Moreover, all $y_n \in [-1,1]$ so that $y' \in [-1,1]$.

Case 1: $x' > 0$. Then $x_n > 0$ for all but finitely many $n$. We can assume w.l.o.g. that all $x_n > 0$, hence $y_n = \sin(1/x_n)$. The continuity of $\sin(1/x)$ on $(0,\infty)$ implies $ y_n \to \sin(1/x')$. By the uniqueness of limits we see that $y' = \sin(1/x')$, hence $(x',y') \in A$.

Case 2: $x' = 0$. Then $(x',y') = (0,y') \in A$.

Concerning your second set $A$ I understand that you mean $y^n = x$ for some $n \in \mathbb{Z}^+$. Clearly $A \subset \mathbb{Q} \times\mathbb{Q}$. If $A$ were open, then $A = int(A) \subset int(\mathbb{Q} \times\mathbb{Q}) = \emptyset$ which is absurd. Next consider any sequence $(y_m) \in \mathbb{Q}$ such that $y_m \to \sqrt{2}$. Then $(y_m,y_m) \in A$, but $(y_m,y_m) \to (\sqrt{2},\sqrt{2}) \notin A$ which shows that $A$ is not closed.

Answer 3

You can show directly that the complement of $A$ is open: let $(x,y)\in\mathbb{R}^2\setminus A$. Case 1. $x<0$: the disc around $(x,y)$ with radius $|x|$ is contained in $\mathbb{R}^2\setminus A$ Case 2. $x=0$: then $y>1$ or $y<-1$; let your radius be $y-1$ or $-1-y$ respectively Case 3. $x>0$: then $y\neq\sin\frac1x$, let $\epsilon=\frac12|y-\sin\frac1x|$ and take $\delta>0$ such that $|z-x|<\delta$ implies $|\sin\frac1z-\sin\frac1x|<\epsilon$. The disc of radius $\min\{\epsilon,\delta\}$ around $(x,y)$ is contained in $\mathbb{R}^2\setminus A$.

As to the second set: $(0,0)$ is in the set but not in its interior: consider points of the form $(r,0)$ with irrational $r$; $(\pi,\pi)$ is in the closure of the set but not in the set itself.