Given a metric Space and four subsets which one is not open?

Question

Given a metric space $\langle \mathbb{R},d \rangle$ where $d$ is metric function defined as $d(x,y) = \begin{cases} \begin{gather*} |x| + |y| & x \not = y \\ 0 & x=y \end{gather*} \end{cases}$

and four subsets $\{A_1,A_2,A_3,A_4\}$ of $\mathbb{R}$ we need to choose one subset that is not open in this space.

$$\begin{cases} \begin{gather*} A_1 & (0,1) \\ A_2 & [-1,4] \\ A_3 & [0,1) \\ A_4 & \{5\} \end{gather*} \end{cases}$$

now according to my textbook definition of an open set - $A$ is an open set if and only if $A = Int(A)$ .
According to Wikipedia - $A$ is open if for every point $x \in A$, $x$ have a neighborhood in $A$.

Now according to any one of these definitions, all four subsets should be not open.
all points of $A_1$ are boundary points since for any $\varepsilon >0$ and $x \in A_1$ any ball $B(x,\varepsilon)$ will contain the point $y = -x \pm \delta$ when $|\delta| < |\varepsilon|$ which in not in $A_1$ (unless $-x+\delta >0$). so $A_1 \not = Int(A_1)$ and hence not open . Same resoning for other three subroups will yield the same conclusions with small diffrence for $A_2$ since $A_2$ actually do contain internal points at the interval $(-1,1)$ but nonetheless $A_2$ have planty of boundary points so $A_n \not = Int(A_n)$ for all 4 options . But according to the question we are to choose only one subset .

What's wrong with my reasoning?

Answer 1

Remember that the definition of an open set $S$ is the following

$\forall x\in S, \exists\epsilon>0$ such that $\{y\;|\;d(x,y)<\epsilon\}\subseteq S$.

Note that if $|x|>0$, if we pick $\epsilon>0$ such that $\epsilon<|x|$, $\{y\;|\;d(x,y)<\epsilon\}=\{x\}\subseteq S$. So, this tells us that sets in this metric that don't contain $0$ are open, since all points that aren't $0$ are interior points in sets they are members of.

Note: There are sets that contain $0$ that also are open in this space. For example, $[-1,4]$ is open (You can show this by analyzing the case of $x=0$: try this as an exercise). But, you can show that only $[0,1)$ is not open.

Answer 2

A $d$-neighbourhood of $0$ looks like a usual neighbourhood: $B_d(x,r) = \{x \in \mathbb{R}: d(0,x) < r\} = \{x \in \mathbb{R}: |x| < r\} = (-r,r)$

A small enough $d$-neighbourhood of $x \neq 0$ is of the form $\{x\}$ as we can take $r < |x|$ and then $d(x,y) = |x| + |y| > r$ for any $y \neq x$. And so $B_r(x) = \{x\}$ for small radii. So all points unequal to $0$ are always interior points of sets they're in.

Now check that this implies only $A_3$ is not open and the others are. $A_2$ contains $0$ and a full neighbourhood of it too. $A_2$ contains $0$ but no points to the left of it. The other are all "isolated point sets".