Discrete metric, singleton open or closed set? Could someone check the following, is my reasoning correct?
EDIT: the following contains errors: see comments

Let $$d_\text{disc}(x,y) = \begin{cases}1 & \text{if } x\not = y\\ 0 & \text{if } x=y\end{cases}$$

Consider a metric space $(M,d_\text{disc})$ and consider $\{ x\} \subset M$. Then:


*

*$\{x\}$ is closed, since $x$ is not an interior point of $\{ x\}$. It is impossible to find a $r>0$ such that $B(x,r) \subseteq \{x\}$,

*$B(x,1)$ is open by definition.


Since in $(M,d_\text{disc})$ the sets $\{ x\}$ and $B(x,1)$ are equal, $\{x\}$ is both open and closed.
Since the union of open sets is open, and the union of closed sets is closed:


*

*$(\forall U \subseteq M)(U$ is open )

*$(\forall U \subseteq M)(U$ is closed )


Supplementary small questions:
Does this imply that the words `open' and 'closed' are not complete? 
If I want to be formal, should I always give the used metric? Writing $d_\text{disc}$-open instead of open?
 A: If $d$ is the discrete metric on any set $X$, then as the union of any open sets in a metric space is open and a set in a metric space is closed if and only if the complement is open, we have: Let $A \subset X$, then $A = \bigcup_{x \in A} B_{d}(x, \frac{1}{2}) = \bigcup_{x \in A} \{x\}$. As $A$ is a union of open sets, $A$ is open for each $A \in \mathcal{P}(X)$.\
It then follows that each set is closed as well.
A: "{x} is closed, since x is not an interior point of {x}. It is impossible to find a r>0 such that B(x,r)⊆{x},"
1)  Closed does NOT mean "not open".  So proving {x} is not open, do not mean {x} is closed.
2)The EXACT OPPOSITE!!!!!
$x$ !!!!!IS!!!! an interior point of $\{x\}$!  If $r = 1$ then $B(x,1) = \{y| d(x,y) < 1\} = \{y|d(x,y) = 0\} = \{x\}\subset \{x\}$.  So $\{x\}$ is open.
I don't know what you mean but "open" and "closed" are "not complete".
Anyway, it's  very well known that in the discrete metric space all sets are both open and closed.
If $S$ is a non-empty set, and $x \in S$. Then $\{y|d(y,x) < 1\} = \{x\} \subset S$ so $x$ is an interior point and all points are interior points in $S$ so S is open.  (The empty set is always vacuously open.).
We use the fact that a set is closed if and only if its complement is open.  So as the complement of all Sets are open (because all sets are open) all sets are closed.
