Let $X = \mathbb R$ with metric $ \rho_0 : X \times X \to \mathbb R$ defined by
$$ \rho_0 (x,y) :=
\begin{cases}
1, & x \ne y;\\
0, & x = y;
\end{cases}
$$
–
$\bullet$ If $A = [0,1)$, then show that $A$ is closed and open in $ \mathbb R$
–
I know that all sets in discrete metric are both open and closed. They are open because if one takes a ball $B_r (x)$ for a $x \in X$ one will invariably find that the only other points contained in the ball besides $x$ are $x$ itself. And we know that singletons are usually open sets. And then the obvious thing is an entire set of singletons will mean that every complement is open. And thus the set is also closed.
But my concern is using the given interval $A$ to show it. I don't think I can visualize this specific interval. Is it the same as saying $A$ is the set $ [ \rho_0 (x,x), \rho_0 (x,y))$
I just need to understand this. I hope you can help!
The next part of question is:
$\bullet$ Produce a set which is not open and closed in $\mathbb R$
And of course I cannot even begin to fathom what this may entail.
Best Answer
The interval $A=[0,1)$ is defined as: $$[0,1):=\{x\in\mathbb{R}\mid 0\leq x<1\}$$ (Which is a subset of $\mathbb{R}$). Any singleton $\{x\}$ is open in the discrete metric, because for instance we can take the ball $B_{1/2}(x)$ to contain only $x$ (not just any $r$ will work). Now since unions of opens are open (axiom in topology), we find (works for any subset $A$): $$A=\cup_{x\in A}\{x\}$$ To be open. Now of course the complement of $A$ will be open by the same reasoning, hence $A$ is closed as well (I think you already understood that but formulated it a bit awkwardly).
For the second question I assume you mean $\mathbb{R}$ with the (standard) Euclidean topology/metric. Then $A$ is not open and not closed: There does not exist an open ball around $0\in A$ which is completely contained in $A$, and the same holds for $1\in\mathbb{R}\backslash A=(-\infty,0)\cup [1,\infty)$.