# Why can't a neighborhood be a finite set?

Rudin defines a neighborhood as follows:

Let $X$ be a metric space endowed with a distance function $d$. A neighborhood of a point $p \in X$ is a set $N_r(p)$ consisting of all $q \in X$ such that $d(p,q) < r$ for some $r > 0$.

He later proves two facts: every neighborhood is an open set, and every finite set is closed.

But it would seem to me that a neighborhood as defined above could be finite. For example, let $X = \{1,2,3\}$ with $d(x,y) = |x-y|$ be a metric space . Consider the neighborhood around $p = 2$ of radius $0.5$, i.e.: the set of all points $q$ in $X$ such that $d(q,p)<0.5$. But the neighborhood is simply $\{2\}$. Therefore, the neighborhood is a finite set, and therefore closed. But every neighborhood is open. This is a contradiction.

So my understanding of a neighborhood is broken somehow. It has further implications: if we define $E = \{2\}$, then the $0.5$-radius neighborhood is $\{2\}$, which is a subset of $E$, which means that $2$ is an interior point of $E$. Since $2$ is the only point in $E$, all points in $E$ are interior points, and $E$ is open. But $E$ is finite, and therefore closed.

I'm probably missing something obvious. Can anyone spot my mistake?

• Nothing about the definition of a set being open or closed precludes a set being both -- merely your intuition. :) Sep 5 '12 at 16:45
• It is possible for are set to be open and closed, as $\{2\}$ in your example. Being open and closed isn't mutually exclusive. Sep 5 '12 at 16:45
• "not closed" is not necessarily "open" and "not open" is not necessarily "closed": "not open" and "not closed" can both be "ajar"! Sep 5 '12 at 16:55
• Obligatory quote: ''…an answer to the mathematician’s riddle: “How is a set different from a door?” should be: “A door must be either open or closed, and cannot be both, while a set can be open, or closed, or both, or neither!”'' - Munkres Sep 5 '12 at 16:58
• Well, in that case, Rudin 2.21 is incorrect, unless there's another hypothesis. Sep 5 '12 at 17:02

"Every finite set is closed" does not imply "every open neighbourhood is infinite".

Let $(X,d)$ be any metric space, and let $F \subseteq X$ be a non-empty finite subset; say $F = \{ x_1, \cdots, x_n \}$ for some $n \in \mathbb{N}$. To prove that $F$ is closed it suffices to prove that $X - F$ is open. So let $p \in X-F$. Then the set $\{ d(p,x_1), \cdots, d(p,x_n) \}$ has a minimum value, say $\delta$, and we know that $\delta > 0$ since $p \not \in F$, and so whenever $d(p,q) < \delta$ we have $q \in X-F$. But this tells you that the (open) ball of radius $\delta$ about $p$ lies in $X-F$, and hence $X-F$ is open.

In your examples, yes, the sets are open and finite, but they are also closed!

The key fact to take home is that sets are allowed to be both open and closed.

However, there is a partial converse: in a dense metric space, every open neighbourhood is infinite.

A metric space $(X,d)$ is dense if for any $x \in X$ and $r > 0$ there exists $y \in X$ with $x \ne y$ and $d(x,y) < r$ $-$ that is, given any point in the space, there are other points which lie arbitrarily close to that point. $\mathbb{R}$ is a dense space, so is $\mathbb{Q}$, so is $(0,1)$, and indeed so is any other non-empty open subset of $\mathbb{R}$.

The fact that open neighbourhoods in dense metric spaces are necessarily infinite is clear from the definition. [It's also clear that a metric space in which every open neighbourhood is infinite is dense, and so the two conditions are equivalent.]

• This makes sense, thanks!
– jme
Sep 5 '12 at 17:11

For an interesting example, let $X$ be any non-empty set, and define $d:X\times X\to\Bbb R$ by $$d(x,y)=\begin{cases}0 & x=y\\1 & x\neq y.\end{cases}$$ This can be shown to be a metric on $X$ (called the discrete metric). One interesting property is that in the metric space $(X,d)$, every subset of $X$ is both open and closed. Even in general metric spaces, there will always be at least two subsets that are both open and closed--namely, the empty set and the whole set.