13
$\begingroup$

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?

$\endgroup$
  • 9
    $\begingroup$ The sets $U_j^x$ are normally required to contain $x$; I consider Krantz's definition defective, though probably inadvertently so. $\endgroup$ – Brian M. Scott Apr 30 '12 at 18:46
  • 4
    $\begingroup$ I would go with accidental omission. $\endgroup$ – copper.hat Apr 30 '12 at 18:48
  • 2
    $\begingroup$ 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). $\endgroup$ – Arturo Magidin Apr 30 '12 at 18:52
  • 2
    $\begingroup$ 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 $y$. $\endgroup$ – rschwieb Apr 30 '12 at 18:53
  • 3
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 30 '12 at 19:08
5
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.