# About interior, closure and etc in $\mathbb{R}^2$

Given this set $A$: $$A =\left\{ \, (x,y) \mid x = 1/n, \ |y| \le n, \ n \in \mathbb{N} \, \right\} \subset \mathbb{R}^2;$$

I'd love to find the interior, closure, set of limit points, set of contour points (I doubt this is a correct English term) and similar things of this nature.

The interior is most likely empty. However it seems that one might need to explicitly add $y$ axis when specifying closure (that is $[A] = A \cup(\{0\} \times\mathbb{R})$ - kinda odd notation) and the same goes for set of limit points. It that correct? Is there some additional trickery?

• The "trickery" is that you need to prove these things formally using their definitions. You can then be sure that there is no additional trickery. – Lord Soth Apr 9 '13 at 21:12
• @LordSoth Fair enough. However I'm not too sure about the proper way to do this hence I would be grateful if somebody could demonstrate a formal proof in question for limit points. – Pranasas Apr 9 '13 at 21:21
• What do you mean by contour point? – Stefan Hamcke Apr 9 '13 at 21:22
• @StefanH. Whose every neighborhood contains both a point of $A$ and a point of $\mathbb{R}^2 \setminus A$. I don't know a proper English word for it. – Pranasas Apr 9 '13 at 21:31
• I don't know if there is an official name. In German it is called Randpunkt which would be boundary point in English. – Stefan Hamcke Apr 9 '13 at 21:39

Let $P=(x,y)$ be a limit point of $A$, i.e. all neighborhoods of $P$ intersects $A\setminus\{P\}$. In other words, there is a (non quasi-constant) sequence in $A$ that converges to $P$ (pick the $n$th element from the disk around $P$ of radius $1/n$, e.g.).
So, assume $A\ni (a_n,b_n)\to (x,y)$. That is, $a_n\to x$ and $b_n\to y$. Now the first coordinate of elements of $A$ form the set $\{1/k\,\mid\,k\in\Bbb N\}$, so, either $a_n\to 1/k$ for some $k$, or $a_n\to 0$. In the former case $a_n$ must be quasi-constant, i.e. $a_n=1/k$ for $n>n_0$ for some $n_0$, and then for these $|b_n|\le k$ must hold, so $(x,y)\in A$ will follow. This shows that the limit points $\subseteq A\cup(\{0\}\times\Bbb R)$.
For the other inclusion: if $(x,y)\in A$, then, one can find a sequence on the line segment of $(x,y)$ that converges there.
Finally, for $(0,r)$, let $N:=\left\lceil\, |r|\,\right\rceil\,\in\Bbb N$, and consider for example the sequence $(a_n,b_n)\in A$ with $$a_n:= \frac1{N+n},\quad\quad b_n:=r\,.$$ This is not quasiconstant and tends to $(0,r)$.