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?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.