Why is a rectangle not a neighborhood of its corners? I'm trying to puzzle out a statement given in the Wikipedia article on topological neighborhoods, which uses this definition:

If $X$ is a topological space and $p$ is a point in $X$, a neighborhood of $p$ is a subset $V$ of $X$, which includes an open set $U$ containing $p$.

The picture used to illustrate the corner argument is this:

I guess I can understand that if $p$ pictured is a member of $V$'s boundary set, which is closed (by the definition of boundary sets?), then the rectangle $V$ does not include an open set $U$ containing $p$, but that's somehow unsatisfying. What if $V$ were actually equal to the whole space $X$? What if $V$ were closed and open at the same time?
More simply, why does the neighborhood of $p$ have to be a disk that extends both inside and outside of $V$?
I have essentially no experience in topology, I'm just doing this out of curiosity.
 A: A neighbourhood of $p$ doesn't actually have to be a disk, but it does have to contain an open set containing $p$.  This means that it will actually contain some open disk centred at $p$, and no matter how small such a disk is, it will not lie completely inside the rectangle $V$.
This example is definitely assuming the ambient space is the Euclidean metric space (not just topological space) $\mathbb{R^2}$ (or at least a big enough piece of it to completely contain the picture!), in order for 'rectangle' to actually have a meaning.  In a general topological space, there is not necessarily such a thing as a rectangle.
You are right that if the ambient space $X$ was only part of $\mathbb{R}^2$, and $p$ was on the edge of $X$, then it could be possible for a rectangle to be a neighbourhood of $p$.
A: If $X$ and $V$ are the same rectangle, then this rectangle is in fact a neighborhood of its corners.  I think you're supposed to assume that the ambient space $X$ is all of $\mathbb{R}^2$ (even though it had to be cropped to a rectangle in the picture.)
Because $\mathbb{R}^2$ is connected, the only subsets that are both closed and open are $\emptyset$ and $\mathbb{R}^2$ itself.
