# Can we live without neighborhood basis but with open neighborhood basis?

I am reading Lee's Introduction to Topological Manifolds, and he declares that neighborhoods always mean open neighborhoods.

So, the definition of a open neighborhood basis goes:

Def Let $X$ be a topological space and $p \in X$. The collection $\mathcal{B}_p$ of neighborhoods of p is called neighborhood basis for $X$ at $p$ if every neighborhood contains some $B \in \mathcal{B}_p$.

This only cares about open neighborhood basis. Is this definition has any defect, I mean is there any mathematical concept that cannot be expressed as an open neighborhood basis, and needs neighborhood basis? If not, why not every mathematician follow this convention?

• This question is rather vague. How can one consider "all the other definitions"? Jun 3, 2015 at 16:48
• I mean, if any defect exists, I think the proposition can be changed without loss of generality.
– Arch
Jun 3, 2015 at 16:59

Let $(X,\tau)$ be a topological space, and $p \in X$.

• an "open neighbourhood" of $p$ is a set $V \in \tau$ such that $p \in V$.
• a "neighbourhood" of $p$ is a set $V \subset X$ such that exists $\Omega \in \tau$, with $x \in \Omega \subset V$.

There's no harm, because:

• every "open neighbourhood" is a "neighbourhood" itself;
• every "neighbourhood" contains an "open neighbourhood".

In many results, it is easier to work assuming that the neighbourhood is already open. So some people prefer to include openness in the definition, to avoid the hassle.

• What he said. I chose to adopt the convention that "neighborhood" always means "open neighborhood," because with the more general definition, there would be more places where I'd have to stipulate "open neighborhood" than there are places where I currently have to stipulate "any set containing a neighborhood." Jun 3, 2015 at 18:57
• Wow. This basically explains everything. Thank you for the answer and the book!
– Arch
Jun 3, 2015 at 19:13
• The general neighbourhood definition allows for statements like: a space is regular iff every point has neighbourhood base consting of closed neighbourhoods, or a space is locally compact iff every point has a compact neighbourhood. These neighourhoods will almost never be open neighbourhoods as well. It's quite handy in many arguments to use the more general one, and just say "open set containing $x$" instead of "open neighbourhood of $x$", which is about equally convenient, I find. Jun 3, 2015 at 19:46
• @Arch: You're welcome! Jun 3, 2015 at 20:00
• @ Henno Brandsma Thank you for the example of locally compactness. Lee defines that: A topological space $X$ is said to be locally compact if for every $p \in X$ there is a compact subset of $X$ containing a neighborhood of $p$.
– Arch
Jun 4, 2015 at 14:37