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

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?

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.

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".