discrete metric, both open and closed. I've checked several answers though, still don't understand last bit.
Taking radius r = 1/2 then every subset is singleton and it is open.
But then how do you deduce it is also closed?
Well, a subset is closed if its complement is open... but then 'every' subset is open hence its complement is empty then closed as empty is both open and closed....??
Thanks.
ps. I haven't learnt about topology space, only metric space.
 A: Note that if every subset is open, then every subset is closed: Given $A \subset X$, then the complement $A^c = X \setminus A$ is a subset, therefore open, and $A^c$ open is equivalent to $A$ is closed.
If you want to be concrete, you can view the complement of a single point as the union of the balls of radius $1/2$ centered on $y$, as $y$ ranges across every point other than $x$.  This union is evidently the complement of $\{x\}$.
A: For every metric space $(X, d)$ and every $x \in X$, the function $y \mapsto d(x,y)$ is continuous. Since the discrete metric is $d(x,y) = \begin{cases} 0, & x = y \\ 1, & x \ne y \end{cases}$, we have that
$$\{x\} = \left\{ y \in X \mid d(x,y) \le \frac 1 2 \right\} = d(x, \cdot)^{-1} \left( \left[ 0, \frac 1 2 \right] \right)$$
which is the preimage of a closed set under a continuous function, hence closed.

Notice that since singletons are open, and since arbitrary unions of open subsets are open, then every subset of $X$ is open: if $S \subseteq X$ then $S = \bigcup _{s \in S} \{s\}$. This means that the topology generated by the discrete metric is the discrete topology, so every subset $S$ is also closed (because its complementary subset is $X \setminus S$ which is open by the previous explanation). In particular, then, singletons are closed.

To conclude, working with the discrete metric on a set is equivalent to working with the discrete topology, in which every subset is both open and closed.
A: let A is subset of discrete metric space.then to show A is closed then A contains all of its limit point.if r=1 clousre of A =A let if c is a limit point which is outside of A let r1 choose such that r1=(min (c,ai_)then open sphere centered with radius r1 not intersect A then A is closed
