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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.

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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 $\{x\}$.

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  • $\begingroup$ 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
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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.

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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

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