Interior, closure, isolated points and boundary of a set of a normed vector space Let $X =(\mathbb{R}^2,||(x_1,x_2)|| := |x_1| +|x_2|)$
be a normed vector space. Find the interior, closure,Isolated points, and boundary of $Y =\{(x, \frac{1}{n})~|~ x\in \mathbb{R} \wedge n\in \mathbb{N}\}$.
I have thought that $open~ball$ of radius $\epsilon$ in this norm about some point say origin looks like a square whose edges are on coordinate axes. Now I don't know how to proceed further. Please help me
 A: First note, that on $\Bbb R^2$ all the norms are equivalent. So instead of using this given norm, one can also consider the natural euclidean norm and get to the same results. This is just a detail, you can solve this problem without knowing this and using the given norm.
Now, assume that $\left( x, \frac 1 n \right) \in Y$ is an interior point of $Y$. Then there exists a $\epsilon > 0$, such that the ball $\Bbb B  \left(\left(x, \frac 1 n \right), \epsilon \right)$ (the open ball around $\left(x, \frac 1 n \right)$ with radius $\epsilon$) is contained in $Y$. But this is not possible, because for each $\epsilon > 0$, the Ball $\Bbb B(x,\epsilon)$ contains a point $(x, y) \in \Bbb B\left( \left(x, \frac 1 n \right), \epsilon\right)$, where $y \in \Bbb R \setminus \Bbb Q$ is a irrational number, so $(x, y) \not\in Y$. This means that the interior of $Y$ is empty, i.e. $\mathring Y = \emptyset$.
Now, let's try to find the closure of $Y$. You can ask yourself the question: Which convergent sequencses $(x_n, y_n) \in Y$ have a limit $(x, y) \not\in Y$? These are exactly the sequences, which converge to a point $(x, 0)$, where $x \in \Bbb R$. So it follows, that $\overline Y = Y \cup \{ (x, 0) \, : \, x \in \Bbb R \}$.
The boundary is now easy to find:
$$ \partial Y = \overline Y \setminus \mathring Y = Y \cup \{ (x,0) \, : \, x \in \Bbb R \} \; .$$
Finally, let's consider the isolated points of $Y$. Take $\left( x, \frac 1 n \right) \in Y$. Then each Ball $\Bbb B\left( \left(x, \frac 1 n \right),  \epsilon \right)$ contains a point $\left( y, \frac 1 n \right) \in Y$, e.g. you can choose $x-\epsilon < y < x + \epsilon$, so $Y$ has no isolated points.
A: Notice that the set $Y$ in $\mathbb R^2$ is actually the set of lines $y= \frac{1}{n}$ ($n \in \mathbb N$).
Now, as $n$ becomes bigger and bigger these lines come closer can closer to the $x$-axis. Thus the answers are:
Interior: Given any point $y$ in $Y$, you can't find a square around $y$ that is completely contained in $Y$. Thus, the interior is empty.
Closure: Notice that, every square around any point on $x$-axis contains some points of $Y$, thus the closure is $Y \cup x$-axis.
Boundary: Closure - Interior, which is the closure itself.
Isolated point: A point is isolated if there exists a square around it disjoint from the set. This can't happen in our case. There are no isolated points.
