# Definition of a neighborhood

• I
Hi,t
I am studying topology at the moment. I have seen that some authors define the neighborhood of a point using inclusion of an open set, while others define the term as open set that contains the point.
In most of the theory I have seen so far, the latter is more convenient to use. Why is there a distinction between the definitions, and what are the advantages of the definition using the inclusion of an open set?

Thank you.

member 587159
I prefer the second definition you mention, because you can formulate the separation axioms in terms of neighborhoods then.

For example, a space is regular if and only if it contains a neighborhood basis of closed sets.

However, I have to admit, that for many purposes, we can work with open neighborhoods, as these form a neighborhood basis for the neighborhoods anyways.

FactChecker
I prefer the second definition you mention, because you can formulate the separation axioms in terms of neighborhoods then.

For example, a space is regular if and only if it contains a neighborhood basis of closed sets.

However, I have to admit, that for many purposes, we can work with open neighborhoods, as these form a neighborhood basis for the neighborhoods anyways.
Thank you for the answer. So why do authors define it using inclusion of open set?

member 587159
Thank you for the answer. So why do authors define it using inclusion of open set?

Personal preference, I suppose.

FactChecker
Gold Member
Can you explain more about the first definition? Do they say that the point is inside the open set that is included? If not, I would think that the point could be on the boundary of a "neighborhood", which would be a problem. If they do specify that the point is inside an open set contained in the neighborhood, then that allows one to call a closed set (or neither open nor closed) a "neighborhood". I think that could be more convenient some times.

Can you explain more about the first definition? Do they say that the point is inside the open set that is included? If not, I would think that the point could be on the boundary of a "neighborhood", which would be a problem. If they do specify that the point is inside an open set contained in the neighborhood, then that allows one to call a closed set (or neither open nor closed) a "neighborhood". I think that could be more convenient some times.
Yes. A set ##S## is a neighborhood of a point ##s\in X## if there is an open set ##U## containing ##s## and is contained in ##S## (##s\in U\subset S##).
Being an open set is independent of being closed. A lot of the theorems are either want a closed or open sets, and neither open nor closed sets are not really interesting or contributing anything (according to my understanding... please correct me if I am wrong). So I am not sure why to add this layer complication.

FactChecker
Gold Member
Yes. A set ##S## is a neighborhood of a point ##s\in X## if there is an open set ##U## containing ##s## and is contained in ##S## (##s\in U\subset S##).
Yes. That is necessary for that definition to be valid.
Being an open set is independent of being closed. A lot of the theorems are either want a closed or open sets, and neither open nor closed sets are not really interesting or contributing anything (according to my understanding... please correct me if I am wrong). So I am not sure why to add this layer complication.
I'm not sure which case you think is a "layer complication". The difference between the definitions is really cosmetic. Their equivalence is easy to prove. Therefore, it can be defined either way and it only takes an easy lemma to use the other, whenever desired.

Last edited:
Yes. That is necessary for that definition to be valid.I'm not sure which case you think is a "layer complication". The difference between the definitions is really cosmetic. Their equivalence is easy to prove. Therefore, it can be defined either way and it only takes an easy lemma to use the other, whenever desired.

By "layer of complication" I meant the definition of including an open set, at least for me, I feel that this is an unnecessary addition (if we define neighborhood to be open).

FactChecker
Gold Member
By "layer of complication" I meant the definition of including an open set, at least for me, I feel that this is an unnecessary addition (if we define neighborhood to be open).
My point is that the two definitions are trivially equivalent. Allowing the freedom of letting a neighborhood not be open can simplify some theorem statements.

mr.tea
mathwonk
Homework Helper
2020 Award
there are open neighborhoods and there are neighborhoods. the point is that all points near your point are in the "neighborhood." no big difference. move on. this is trivial.

member 587159
WWGD
Gold Member
To be pretentious, this condition is called normality ( one of the separation axioms), which goes, I think, without being T1. And metric spaces are normal.

member 587159
To be pretentious, this condition is called normality ( one of the separation axioms), which goes, I think, without being T1. And metric spaces are normal.

I'm curious about what condition you are talking? Normality is that every two disjoint closed sets can be separated by disjoint neighborhoods.

WWGD