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.
 A: In metric spaces, open balls may have just one point: Let $X$ be any (non-empty) set and consider the following metric in $X$: $d(x,y)=\begin{cases}0&,if\ x=y\\1&,if\ x\neq y\end{cases}$. Then the open ball of radius $1/2$ centered at any $x\in X$ is simply $\left\{x\right\}$. For a less artificial example, take $Y=[0,1]\cup\left\{17\right\}$ with the metric induced from $\mathbb{R}$. Then the open ball (in $Y$!) centered at $17$ with radius $4$ is just $\left\{17\right\}$.
On the other hand, if you have a non-trivial normed space $X$ (over $\mathbb{R}$, say), then any open ball in $X$ will have infinite points.
A: 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 $X$.
A: It depends on the space. In $\mathbb{N}$ with the usual topology inherited from the real line, the open 1-ball about any point is just that point. Such points are said to be isolated.
Your example is correct. It is a trivial space consisting of a single point, and you have observed the only topology that exists for that space.
