Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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 $x$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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