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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:

enter image description here

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.

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If the rectangle is the whole space, then the part of the open disk in the picture that is outside the rectangle is empty. – André Nicolas Feb 22 '13 at 22:01
You're correct that if $V = X$ is the entire ambient space, then it is a neighborhood of its corners. The example is assuming that the rectangle is being considered as a subset of the entire plane. – MartianInvader Feb 22 '13 at 22:35
Glad to know my intuition works a little bit. Thanks. – Trevor Alexander Feb 23 '13 at 5:36
up vote 2 down vote accepted

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 $p$.

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I'll take a look at metric spaces then and try to put it together. Thanks. – Trevor Alexander Feb 22 '13 at 22:11
Sorry, I had made a mistake in the first sentence. A neighbourhood needn't be an open set. I've fixed it now. You don't really need anything about metric spaces, since the topology on $\mathbb{R}^n$ as a metric space is the same as the usual topology (basic open sets are open balls). – Tara B Feb 22 '13 at 22:14
"Note that due to the definition of the norm, the unit circle is always convex and centrally symmetric (therefore, for example, the unit ball may be a rectangle but cannot be a triangle)." ( By analogy, do the symmetry and convexity requirement of distance functions give rise to the 'ball-like' shape of neighborhoods in metric spaces? – Trevor Alexander Feb 22 '13 at 22:17
Where is that from? – Tara B Feb 22 '13 at 22:18
Specifically, there is the sentence: "Note that due to the definition of the norm, the unit circle is always convex and centrally symmetric (therefore, for example, the unit ball may be a rectangle but cannot be a triangle)." If the norm is a distance function on a vector space like $\mathbb{R}^2$, this is intuitively satisfying to me, and fits with the image on the neighborhood article. – Trevor Alexander Feb 22 '13 at 22:22

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.

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So the image is kind of an appeal to intuition based on the assumption that $X$ is part of $\mathbb{R }^2$? – Trevor Alexander Feb 22 '13 at 22:10
I suppose images are always appeals to intuition. In any case we have to assume $X \subseteq \mathbb{R}^2$ to make sense of the word "rectangle" in the example. No boundary of $X$ is drawn, so I'd assume $X$ is all of $\mathbb{R}^2$. – Trevor Wilson Feb 22 '13 at 22:11

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