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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?

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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
I would go with accidental omission. – copper.hat Apr 30 '12 at 18:48
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
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
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 Answer 1

The super script $x$ is to indicate that those are open sets that containing $x$. Thus it follow immediately that the neighbourhoods do in fact contain $x$

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While I agree with you in other contexts, this is not really what the question was about. It does not follow at all from the definition I provided that the neighbourhoods should contain $x$. Arturo Magidin's comments above have answered the question. – roo Aug 9 '14 at 8:25

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