Can an open ball have just one point. As per my understanding it cannot. Please clarify. I am new to functional analysis and am just learning. To my understanding an open ball must have at least 2 points else its definition will not be satisfied.
Now if I have just an empty set and this open ball, why cannot it constitute a Topology.
I see it satisfying intersection and union conditions. 
I agree that the question is too basic!
Definition of open ball : $$B_r(x)= \{y \in E | d(x,y)<r \}$$
$(E,d)$ is the metric space.

You say that you are studying functional analysis, so perhaps you are mainly interested in Banach spaces. But even in that context, you are only almost correct - any open ball of a non-trivial Banach space is infinite. The trivial Banach space $V=\{0\}$ consists of just its zero vector $0$, and thus for any $r>0$,
$$B_r(0)=\{v\in V:|v-0|<r\}=\{v\in \{0\}:|v|<r\}=\begin{cases}
\{0\}&\text{if }|0|<r,\\
\varnothing&\text{if }|0|\geq r
\end{cases}=\{0\}=V,$$
which is therefore both an open ball and a singleton set.
Of course, most metric spaces are not Banach spaces, and there are many other counterexamples. Basically, for any metric space $X$, if $x\in X$ is an isolated point, then $\{x\}$ will be an open ball of $X$.
