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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
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  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 $(-1,1)$.

    A basis for the neighborhood system is a collection of neighborhoods with the property that any neighborhood contains (at least) one element of the system. They can be thought of as "small enough representatives" so that at least one of these representatives is contained in any given neighborhood.

  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.)

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I think you identified the source of my confusion. Thank you very much. –  Indrid Cold May 1 '12 at 22:31
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