14

In Steven G. Krantz' A Guide To Topology, a countable neighborhood base is defined:

Let $(X,U)$ be a topological space. We say that a point $x\in X$ has a countable neighborhood base at $x$ if there is a countable collection $\{U_{j}^{x}\}_{j=1}^{\infty}$ of open subsets of $X$ such that every neighborhood $W$ of $x$ contains some $U_{j}^{x}$.

Here is a link.

Now, to define a neighborhood base at $x$, the obvious thing to do would be to simply drop the countable requirement, and replace $\{U_{j}^{x}\}_{j=1}^{\infty}$ with some collection $\{U_{\alpha}^{x}\}_{\alpha\in J}$ with index set $J$.

My question: (and I've seen this same definition in other books) Why do we not require that each $U_{\alpha}^{x}$ contain the point $x$? My intuitive idea of what a neighborhood base ought to be completely falls apart without this requirement. All examples of neighborhood bases seem to satisfy this. Is it a consequence of the definition?

Is there a better way to think about neighborhood bases?

Norbert
  • 56,803
roo
  • 5,598
  • 9
    The sets $U_j^x$ are normally required to contain $x$; I consider Krantz's definition defective, though probably inadvertently so. – Brian M. Scott Apr 30 '12 at 18:46
  • 4
    I would go with accidental omission. – copper.hat Apr 30 '12 at 18:48
  • 3
    Petersen defines a neighborhood basis for $x$ as a system $\rho$ of subsets of $X$ such that for all neighborhoods $A$ of $x$ there is a $B$ in $\rho$ that is a neighborhood of $x$ and such that $B\subseteq A$. So the elements of the basis are not required to contain $x$, but the witnesses are. ("Analysis Now", Gert K. Petersen, GTM 118, page 10). – Arturo Magidin Apr 30 '12 at 18:52
  • 4
    The way to think about a neighborhood base at $x$, as you probably already do, is to think of them as a special subcollection of neighborhoods of $x$ which get inside of all neighborhoods of $x$. For example, in $\mathbb{R}$ with the metric topology, the balls of rational radius around $x$ form a (countable) base at $x$. In a topological space with an isolated point $y$, the set ${{y}}$ is a neighborhood base for $y$. – rschwieb Apr 30 '12 at 18:53
  • Thanks for clarifying everyone. – roo Apr 30 '12 at 18:54
  • 4
    For an example of a system that satisfies the given definition but which probably "should not" be a neighborhood base, take the Sierpinsky space $X={a,b}$, $\tau={\varnothing, {a}, {a,b}}$, the point $b$, and the set ${a}$ as the only element of your basis. Then every neighborhood of $b$ contains an element of our neighborhood basis, but we probably don't want to call ${{a}}$ a "neighborhood basis for $b$". Notice that this does not satisfy the definition in Petersen. Under Petersen's definition, sets that don't contain $x$ don't really matter. – Arturo Magidin Apr 30 '12 at 19:08

1 Answers1

6

As was mentioned by a couple people in the comments, the answer is that this is neither the usual nor a particularly good definition of a neighborhood base.

In fact, let $X$ be any topological space, let $x\in X$, and define $\mathcal{B}_x:=\{ \emptyset \}$. Then certainly, for any neighborhood $W$ of $x$, there is some element of $\mathcal{B}_x$ which is a subset of $W$. This, however, is silly.

Suppose, however, that we even amended the definition you gave to require that elements of a neighborhood base were all nonempty. Even in this case, the definition is lacking.

One way to see this is because, for the usual definition, we have that, if each $\mathcal{B}_x$ is a neighborhood base at $x$ for all $x\in X$, then $U\subseteq X$ is open iff for every $x\in U$ there is some $B_x\in \mathcal{B}_x$ such that $B_x\subseteq U$. With even the amended definition of what you gave (which hereafter I shall refer to simply as "your definition"), this is no longer so.

For example, take $X:=\{ 0,1,2\}$ with the topology $\{ \emptyset ,\{ 0\} ,X\}$, and define $\mathcal{B}_0,\mathcal{B}_1,\mathcal{B}_2$ all to be $\{ \{ 0\} \}$. Then, each $\mathcal{B}_x$ is a neighborhood base according to your definition. But now it is the case that for every element $x\in U:=\{ 0,1\}$ there is some element $B_x\in \mathcal{B}_x$ such that $B_x\subseteq U$, even though $U$ is not open!

Jonathan Gleason
  • 7,861