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....??


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 $\{x\}$.

  • $\begingroup$ Is 'A' any subset? Sorry but I don't get why it's $A^c is open. $\endgroup$ – mathstock Apr 22 '15 at 20:48
  • $\begingroup$ 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! $\endgroup$ – Rolf Hoyer Apr 22 '15 at 20:50
  • $\begingroup$ The answers to this question might be helpful if you need elaboration $\endgroup$ – Rolf Hoyer Apr 22 '15 at 20:57
  • $\begingroup$ 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? $\endgroup$ – mathstock Apr 22 '15 at 21:05
  • $\begingroup$ Let $B = X\setminus A$. All subsets of $X$ are open, and $B$ is a subset of $X$, so $B$ is open. $\endgroup$ – 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.

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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.