Closure, interior, and boundary of a subset Consider $\Bbb R^2$ with standard distance and a subset $A= \left([0,1] \times [1, \infty ) \right) \cup \bigcup _{n=2}^ \infty \{(x, \frac1n):0 \le x \le \frac1n \}$
Question: Is A closed? Find its closure, interior, and boundary.
My answer is that A is not closed, although I'm not entirely sure. My reasoning is that the complement of A in $\Bbb R^2$ contains the top side of the "rectangle" at infinity, i.e. $\{ (x, \infty) : 0 \le x \le 1 \}$, which is not open. (what I'm not sure about is if $\{ (x, \infty) \}$ is an actual "point" that "exists" in $\Bbb R^2$...)
So for the closure of A, I get that it's $([0,1] \times [1, \infty])$ plus the other bits (the horizontal segments below the rectangle). The interior of A is $(0,1) \times (1, \infty)$, without the other bits since they have no interior. The boundary is the 4 sides of the rectangle (including the one at infinity) plus all the line segments below it, namely $\{(0,y) \cup (1,y): 1 \le y \le \infty \} \cup \{(x,1) \cup (x, \infty): 0 \le x \le1 \} \cup \bigcup _{n=2}^ \infty \{(x, \frac1n):0 \le x \le \frac1n \}$
I am counting on someone to point out my mistakes, of which there may be plenty.
Thanks!
 A: After more investigation, it seems $A$ is not closed. We can find a limit point of $A$ not contained in $A$. Consider the point $(0,0)$. We know For all $\varepsilon >0$, the open ball $B_\varepsilon\left((0,0)\right)$ must contain at least one point in $$\bigcup _{n=2}^\infty \left\{\left(x, \frac{1}{n}\right):0 \le x \le \frac{1}{n}\right\}$$ as there exists $N \in \Bbb{N}$ such that $\frac{1}{N} < \varepsilon$ and hence $$B_\varepsilon((0,0)) \bigcap \left[\left\{\left(x, \frac{1}{N}\right):0 \le x \le \frac{1}{N}\right\}\setminus \{(0,0) \}\right]\neq \emptyset$$ as the intersection above contains the point $\left(0,\frac{1}{N}\right)$. This demonstrates that $(0,0)$ is a limit point of $A$ that is not contained in $A$. We now know the closure of $A$ must include $(0,0)$. I believe you correctly identified the interior of $A$. To be sure of the boundary of $A$, you can use the definition $\text{Bd}(A) = \overline{A} \setminus \text{Int}(A)$ since you know $\overline{A}$ and $\text{Int}(A)$.
A: The "infinite" points are not true points of $\mathbb R^2$, hence they do not enter here. The rectangle is already closed, but the "other bits" give the trouble. The easiest way to understand closures in $\mathbb R^k$ is to see adherent points as limits of points in the set. In our case, the set contains the points $p_n=(0,\frac{1}{n}), n\ge2$. The limit of this sequence is clearly the origin $p_0=(0,0)$, which is not in $A$. Hence the origin is a point in $\overline A\setminus A$ and $A$ is not closed. In fact, this is the only missing point, that is, $\overline A=A\cup\{(0,0)\}$. For instance, the points in the $y$-axis below $1$ are only 
$p_0,p_1,p_2,\dots$. Note that given any $(0,y)$ between two of these, there is always a small disc around that does not touch $A$. 
