Are open disks open in $\mathbb{R}^3$? Consider $\mathbb{R}^3$ with the topology generated by open 3-dimensional balls $B_{a, \epsilon} := \{x\in\mathbb{R}^3\ |\ d(a, x)<\epsilon\}$, where $d(a,x)$ is the Euclidean distance between $a$ and $x$. Are "open" disks (ie. 2-dimensional balls without boundary) open in $\mathbb{R}^3$?
I don't think a 2-ball without boundary can be constructed as a union of open 3-balls (unless the 3-balls have 0 radius, but then they're points, and then points are open and we degenerate into the discrete topology, so this doesn't work). Maybe as an intersection?
In general, are $k$-balls without boundary open in $\mathbb{R}^n$ with the $n$-ball topology (for $k<n$)?
 A: No, they are not since if $x$ is a point of such a ball $B_k, k<n$, the $n$-ball $B_n(x,c)$ is not contained in $B$ for any $c>0$.
An open ball whose radius is zero is empty.
A: No, open discs of dimension $k< n$ are not open in $\Bbb R^n$, although it is a very good question. To see this, recall that a set is open iff it is equal to it's interior. Now, for $x\in \Bbb D^k$, $x$ must be of the form $(x_1,\ldots, x_k,0,\ldots,0)$. Thus for all $\epsilon >0$, the point $(x_1,\ldots, x_k, \frac{\epsilon}{2},0,\ldots, 0)\in B_\epsilon(x)$, so $x$ cannot be in the interior of $\Bbb D^k$.
A: No, they're not. When we're talking about open balls in $n$-space, we require the points to extend radially outward in all directions. So for $k<n$, $k$-balls are lacking dimensions, so to speak. They are nowhere dense in $\mathbb{R}^n$, in fact. 
A: No, they're not open. An easier way to think about it is "are open balls in $\mathbb{R}$ open in $\mathbb{R^2}$?"
An open ball in $\mathbb{R}$ is an open interval. In $\mathbb{R}$, every point in an open interval is an interior point. This is easy to show by constructing some open interval $(c-\epsilon, c+\epsilon)\subset (a,b)$ where $c \in (a, b)$.
In $\mathbb{R}^2$ however, every point in the interval is a boundary point. If you take some point $c$, then for each $\epsilon$ there will be a point in $B_\epsilon(c)$ which is not in the interval. On the other hand, $c$ is definitely in the interval. Thus each open ball at $c$ contains points both in and out of the interval, so every point in the interval is a boundary point, and there is no interior. The ball in $\mathbb{R}$ is not open in $\mathbb{R}^2$.
Using exactly the same argument, any open ball in $\mathbb{R}^n$ is not open in $\mathbb{R}^{n+1}$.
