One question about open sets in topology Here is one question given by my topology teacher. But I feel a little bit puzzled.

Given a metric space $(X,d), U⊆X$ is open if and only if for any $x\in X$,then there exist $\epsilon>0$, s.t. $B(x,\epsilon)⊂U$, where $B(x,\epsilon)=\{y\mid y \in X \text{ and } \space(x,y)<\epsilon\}$.

My question is: In Mathematical Anaylysis, we define an open set as all the points in it are interior points, and I find that second part of the question ("for any $x\in X$,then there exists
$\epsilon>0$, s.t. $B(x,\epsilon)⊂U$, $B(x,\epsilon)=\{y\mid y \in X \text{ and }
\space(x,y)<\epsilon\}$") is telling exactly the same story, so since we define an open set as including all its interior points, how can we prove this definition? Thanks in advanced.
 A: In my Topology class, I  learned two definitions of a topological space. One is the Neighbourhoods definition and the other is the Open sets definition. In the neighbourhoods definition, we call a subset $O$ of a topological space $X$ open if it is a neighbourhood of each of its points. This leads us to the properties of open sets we are familiar with-
(1) Union of any collection of open sets is open
and
(2) Intersection of any finite number of open sets is open
In the Open sets definiton, we work the other way around, that is, we start with the idea of an open set and then we build up a collection of neighbourhoods for each point. We have a set $X$, and we have a nonempty collection of subsets of it, which we call open sets and these sets have properties (1) and (2). For a point $x\in X$, a subset $N$ of $x$ is called a neighbourhood, if we can find an open set $O$ such that $x\in O⊆ N$. This definition of neighbourhood makes $X$ into a topological space, by the neighbourhood axioms.
These two notions of 'openness' and 'neigbourhoodness' coincide, that's why we rephrase our defnition of toplogical space in terms of open sets because then we have fewer axioms to deal with.
In our class, for a metric space, we defined a set to be open in the way it is written in your statement above. So you see the problem here, you have to tell us first what your original notion of an open set is. 
