# 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 find it difficult to imagine what this space looks like. I think it consists of all sequences containing ones and zeros.

Now in order to prove that every subset is open, my books says that for $A \subset X$, $A$ is open if $\,\forall x \in A,\,\exists\, \epsilon > 0$ such that $B_\epsilon(x) \subset A$.

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

• 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.

• Shouldn't it be that "Since all sets are open, their complements are closed"? – Khallil Jan 24 '16 at 21:44
• @Khallil that is also true, by the definition of 'closed'. But the complements are open as well here, since all sets are open. My point was to make the transition from "all sets are open" to "all sets are complements of open sets, and therefore closed as well". – Robert Mastragostino Jan 28 '16 at 15:35
• @Robert Mastragostino Any set E in a discrete metric does not have a boundary point and so has both the properties hold true:1.The set E contains all its boundary points 2. The set E does not contain any of its boundary points. Hence the set E is simultaneously open and closed.Can one argue like this? – nrynn May 31 '17 at 13:57
• For visualization purposes: If I understand your radius in a ball correctly, the point at the center is from where all connections must originate. The radius is of fixed length, it is the distance from the center to any point on the surface of a perfect sphere. The points on the surface can be anywhere on the sphere, which is the open part of the formula, but for every point on the surface that connects via the radius to the center, the point where the center is must be the same. This fixed point is what allows you to use ratios involved in formulas involving a radius within a sphere. – G_V May 25 '18 at 9:06
• Would it be possible to have a ball with radius bigger than or equal to 1 with this metric space? – Aqqqq Sep 29 '18 at 6:55

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.

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.

Let $$(X,d)$$ be a metric space. Suppose $$A \subset X$$. Let $$x\in A$$ be arbitrary. Setting $$r = \frac{1}{2}$$ then if $$a \in B(x,r)$$ we have $$d(a,x) < \frac{1}{2}$$ which implies that $$a=x$$ and so a is in A. (1)

To show that A is closed. It suffices to note that the complement of A is a subset of X and by (1), it is open hence A must be closed.