# Show that in a discrete metric space, every subset is both open and closed.

I need to prove that in a discrete metric space, every subset is both open and closed. Now, I have problem imagine how this space looks like. I think it contains of all sequences containing ones and zeros.

Now in order to prove that every subset is open, my books says that

$A \subset X$

$A$ is open if $\,\forall x \in A,\,\exists\, \epsilon > 0$ then $B_\epsilon(x) \subset A$

I was thinking that since A also will contain only zeros and ones, it must be open. Could someone help me ironing out the details? =)

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Re: "I think [this space] contains of all sequences containing ones and zeros": No, that's not what "discrete metric space" is. A discrete metric space is any set plus the discrete metric. –  ruakh Sep 11 '12 at 18:18

The discrete metric just says that $$d(x,x)=0$$ $$d(x,y)=1,\ x\neq y$$

So say your ball has radius $r$. If $r<1$ then the only point it contains is the point it's centred on. So any single point has a ball of some radius around it containing only that point. This is the same thing as $B_{0<r<1}(x)=\{x\}$, so we know that every singleton is open. And now we're actually done! Since now we know that any point $x$ in a set $A$ has a ball containing it, because we can always construct a ball that only contains $x$! Since all sets are open, their complements are open as well. This implies that all sets are also closed.

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In a discrete space, take $\epsilon=\frac{1}{2}$. Then the ball contains only the point itself so that it is a subset of $A$. Every subset is closed because the complement of an open set is closed.

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I suggest a new outlook - I think of a discrete metric space as a collection of particles, all spread apart from each other. The particles are the points of the space, and "spread apart" means that $d(x,y)\geq D>0$ for any different particles $x$ and $y$. Usually the definition of a discrete metric space is that the distance between any two points is $1$, but this is just a model (and in fact any metric space with $d(x,y)\geq D>0$ for all $x\ne y$ is homeomorphic to the discrete metric space on the same number of points, and we're only interested in topological things so it doesn't matter that they aren't isometric).

Now it should be clear how, for each point, to take an open ball that consists only of that point, proving that points are open. Will's answer then gives some hints about how this implies the result you want.

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In a discrete topological space, all subsets are open by definition. Consider a fixed family B of subsets of the overall space X. Since all subsets the universal set X are open, their complements relative to the set are closed. But it's not hard to show that the complements of the complements of each set is the family B itself! So every set in B is also closed. So every subset of X is either open or closed.

Now the trick is to formulate this argument specifically for metric spaces in terms of open balls. Sketch of Proof: Formulate the family B as a nested sequence of open balls of equal constant distance c centered at an arbitrary point x Note c doesn't have to be 1,but make c=1 if it makes it easier for you. Demonstrate that in addition to being closed, the set of complements of open balls in B is itself a sequence of open balls in X. (This is a perfectly general argument since each metric space has a countable neighborhood basis of each point(local base),but you don't have to worry about this technicality for this argument.) Consider the possible values of c in the construction of the open ball sequence in comp(B). This will show each subset of comp(B) is either a singleton (which is open) or an open ball containing 2 points of diameter c. This will give you the proof.

Be careful!

Addendum:Clearly what I meant was that each metric space was first countable. Adam Smith,who loves to cherry pick my posts,pointed this out. I've corrected it and it was a fairly egregious error.

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-1. I have no idea what most of this means, and it contains several false statements (eg it is not true that every metric space has a countable basis; think of a nonseparable Banach space). Even if it can be made correct, it is much more complicated than the answers that were posted long before you posted yours. –  Adam Smith Sep 11 '12 at 20:03
@Adam Firstly,clearly what I meant was that each metric space was first countable.I've corrected it-it WAS a fairly egregious error. Secondly,I think it's fairly clear if you follow it carefully-this is just a sketch and isn't meant to be a rigorous argument-I want the OP to prove it for himself/herself.Lastly-I hadn't seen Robert and Will's answers when I wrote this-they basically gave the full answer I was merely trying to sketch for the OP. –  Mathemagician1234 Sep 12 '12 at 2:55