# 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.

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 $\{x\}$.
• Is 'A' any subset? Sorry but I don't get why it's $A^c is open. – mathstock Apr 22 '15 at 20:48 • The point is that any subset is open. To check this, you need to show that for any$y\in A^c$that there is some open set$U$such that$y\in U \subset A^c$. Choose$U = \{y\}$, the ball of radius 1/2 around$y$, done! – Rolf Hoyer Apr 22 '15 at 20:50 • The answers to this question might be helpful if you need elaboration – Rolf Hoyer Apr 22 '15 at 20:57 • Thanks for your effort. Actually, I've already seen that post. What I'm confused is that 'Since all sets are open, their complements are open as well.' What I thought is, open if its complement is closed. closed if its complement is open. A- open set then X∖A is closed which is complement of A. How is this open? – mathstock Apr 22 '15 at 21:05 • Let$B = X\setminus A$. All subsets of$X$are open, and$B$is a subset of$X$, so$B$is open. – Rolf Hoyer Apr 22 '15 at 21:13 For every metric space$(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.