# Definition of Basis for the Neighborhood System

I'm trying to learn a bit about topology through independent study. I've been using Bert Mendelson's "Introduction to Topology - 3rd edition". I'm having a lot of fun but I'm a bit confused regarding definition 4.9 on pg 45. I will reproduce it hereafter:

Definition 4.9 - Let $a$ be a point in metric space $X$. A collection of neighborhoods $\mathcal{B}_a$ is called a basis for the neighborhood system at $a$ if every neighborhood $N$ of $a$ contains some element $B$ in $\mathcal{B}_a$.

Here is the source of my confusion, please correct me if I am wrong:

1) Every neighborhood of $a$ must contain $a$ it self, so shouldn’t any neighborhood of $a$ be automatically a basis of the neighborhood system at $a$?

2) If this is true (and I'm hoping it is false) what is the condition that forces $\mathcal{B}_a$ to grow beyond a trivial set such as $\mathcal{B}_a = \{a\}$?

Thank you.

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The members of $\mathcal{B}_a$ are required to be nbhds of $a$; $\{a\}$ isn't a a nbhd of $a$ unless $a$ is an isolated point of $X$, in which case $\{\{a\}\}$ is a nbhd base at $a$. – Brian M. Scott May 1 '12 at 21:38

1. Yes, every neighborhood of $a$ must contain $a$; but it is not true that any particular neighborhood is a basis for the neighborhood system, because it may not be contained in every neighborhood. For example, consider the real line and $a=0$. The set $(-1,1)$ is a neighborhood of $a$, but is not by itself basis of the neighborhood system at $0$ because, for example, the neighborhood $(-1/3,1/2)$ does not contain the set $(-1,1)$.
2. First, note that $\mathcal{B}_a$ is a set of sets, not a set of elements of the space. Perhaps you mean $\mathcal{B}_a=\bigl\{\{a\}\bigr\}$, rather than $\mathcal{B}_a=\{a\}$. Second: if $\{a\}$ is a neighborhood of $a$ (it may not be: for example, it may not be open!) then it is true that one can take $\mathcal{B}_a$ to be just $\bigl\{\{a\}\bigr\}$. But for example, $\{\{0\}\}$ is not a basis of the neighborhood system for $0$ on the real line (with the usual topology), because $\{0\}$ is not even a neighborhood of $0$.
It is true that if there is a "smallest open set that contains $a$", then one can take that set alone to be a basis of the neighborhood system. But in many topological spaces, no such set exists. In the real numbers with the usual topology, there is no "smallest open set that contains $a$", so a basis of the neighborhood system at $a$ needs to have more than one element: given any open set that contains $a$, there is always a strictly smaller open set that contains $a$, so a basis for the neighborhood system in this topological space would necessarily have to contain infinitely many elements.
(An example of a basis of the neighborhood system at $a$ for the real numbers with their usual topology would be $$\mathcal{B}_a = \left\{ \left.\left(a -\frac{1}{n},a+\frac{1}{n}\right)\right|\, n\in\mathbb{N}\right\}$$ which you can verify: each element of $\mathcal{B}_a$ is a neighborhood of $a$, and every neighborhood of $a$ contains at least one element of $\mathcal{B}_a$; you can also replace the sets in $\mathcal{B}_a$ with the corresponding closed intervals, which are also neighborhoods and have the relevant properties.)