Show that the discrete topology on $X$ is induced by the discrete metric Let $X$ be a set. Show that the discrete topology on $X$ is induced by the metric
$d(x, y) =
\left\{
 \begin{array}{ll}
  1  & \mbox{if } x \neq y \\
  0 & \mbox{if } x = y
 \end{array}
\right.$

Unfortunately, I don't understand this at all! By discrete topology we mean that every element of X must alone be in an open set, although every possible gathering of any elements with others also form an open set. In a metric, by definition (?) $d(x,y)=0$ if $x=y$; thus, doesn't matter if it induces discrete topology or any other topology, "$d(x,y)=0$ if $x=y$" holds trivially. But why 'distance' between some element to ANY other element is $1$?
Thank you.     
 A: A metric $d$ on a set $X$ induces the topology $\mathcal{T}$ by taking the open balls $B(x,r) = \{y \ : \ d(x,y)<r\}$ as basic open sets. What you need to show is that the $d$ you are given produces the topology $\mathcal{T}$ which is the same one as the discrete one.
In the definition of $d$ the $x=y$ case is included precisely because the definition requires that $d(x,x)=0$. It would be a good exercise to show that this function $d$ is actually a metric.
Now, all we need to do is show that a subset $Y$ of $X$ is open in the discrete topology iff it is a union of open balls. As you noted, every $Y\subseteq X$ is open in the discrete topology, so, all we need to show is that every $Y$ is a union of open balls. I'll get you started. If $Y = \{x\}$ then $Y = B(x,1/2)$ (you should check this). Now, can you show that an arbitrary subset of $X$ is a union of open balls?
A: For any $x$ in $X$, $S(x,1)= \{y:d(x,y)<1\}$
But according to definition,
$S(x,1)=\{x\}$ which is open in the discrete topology .
So we can conclude the result
