Question about open subsets in topology I am having trouble understanding the following theorem.
Any open ball B(x;r) in a metric space X is an open subset of X.
I understand that intuitively it contains all the points in the open ball except the boundary points. But for this ball to be an open subset, every point of B(x;r) must be an interior point of B(x;r) right?
But for a point to be an interior point,  there needs to be a ball with the point being the center. The question is that there is always a tiny bit of a distance between this center and the boundary of B(x;r) and within this distance(gap), there exist points of B(x;r) which also have to work as a center point right? Doesn't that mean that not all points of B(x;r) are interior points? 
To put it another way, pick a point that is extremely close to the bounday of B(x;r). Can you make an open ball inside this B(x;r) with that point being the center no matter how close the point is to the boundary?
 A: Yes, logically (rather than empirically) it should be ensured. To see this, let $\delta \geq 0$; let $r - \delta \leq |y-x| < r$. Then you can make $y$ as close to the imaginary boundary of $B(x;r)$ as desired by making $\delta$ sufficiently close to $r$. Now can we form an open ball $B(y; r')$ of center $y$ and radius $r' > 0$? Yes of course; for example, taking $r' :=r - (\delta/2)$ suffices.
A: If $(X,d)$ is a metric space, then for all $x \in X$ and $\epsilon >0$, the set $B(x,\epsilon)$ is open. 
To see why this is true, suppose that $y \in B(x,\epsilon)$. Then we have that $d(x,y) < \epsilon$. Thus set $r=\epsilon - d(x,y)>0$. If $z \ in B(y,r)$, then $d(z,y) < r$.
From the triangle inequality, we then have that $d(x,z) \le d(x,y) + d(y,z) < d(x,y) + r = d(x,y) + \epsilon - d(x,y) = \epsilon$. Thus $ z \in B(x,\epsilon)$ and $B(y,r)$ is contained in $B(x,\epsilon)$.
So no matter how close a point $y \in B(x, \epsilon)$ is to the boundary (ie how close $d(x,y)$ is to $\epsilon$) we can always find a really small positive value for $r$ such that $B(y,r)$ is contained in $B(x,\epsilon)$.
A: It's wrong as written (because in trivial topology where only open sets are empty one and whole space, not surprisingly, not all balls are open) and trivial in corrected version (every metric space  $X$ posess metric topology, where you explicitly state the open subsets are arbitrary unions of finite intersections of open balls $B(x_0, R) : x \in X, d(x_0, x) < R$). So, if you translate whatever strange definitions you have to standard ones, it's not a theorem but tautologically true or false statement.
