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Ask another question about Neighborhood(Ne):

There are two definitions:

  1. http://en.wikipedia.org/wiki/Topological_space (Neighbourhoods definition)
    That is:
    (a) if N ∈ N(x), then x ∈ N
    (b) if N ∈ N(x), N⊆M, then M ∈ N(x)
    (c) if N1, N2 ∈ N(x), then N1 ∩ N2 ∈ N(x)
    ...
  2. http://en.wikipedia.org/wiki/Neighbourhood_(mathematics)
    That is:
    If X is a topological space and p is a point in X, a neighbourhood of p is a subset V of X that includes an open set U containing p.

My question is, in 2., it requires an open set U; however, in 1., it seems that it does not require an open set. Why? Does two definitions relate each other?

Asaf Karagila
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sleeve chen
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  • This looks suspiciously close to http://math.stackexchange.com/questions/922308/def-about-neighborhood-on-wiki – Asaf Karagila Sep 07 '14 at 15:25
  • Not exactly, the previous one I ask is all from 2., this question is from two websites. In fact, I watch a lecture on youtube, the professor introduced the definition of 1., it does not mention 2. – sleeve chen Sep 07 '14 at 15:30
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    Definition 1 seems to be more common in the French literature, while Definition 2 - in the US. – Moishe Kohan Sep 07 '14 at 15:46

2 Answers2

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You can build a topology either on top of the notion of neighborhood or on top of another notions (open sets, closed sets, or closures). Each of the axiomatizations should be equivalent to another.

Apparently, the first definition is an axiomatization of a topological space done via neigborhood (see the wiki article).

The second definition is actually a definition of neighborhood in a topological space axiomatized using open sets.

Jindra Helcl
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Given a topological space $X$, I may define for every $x\in X$ a family of subsets of $X$ called neighbouroods of $x$ by saying that $N\in N(x)$ if $N$ contains an open set which contains $x$.

On the other hand, if I have a set $X$ and a family of subsets $N(x)$ for every $x\in X$ with the properties

  • if $N\in N(x)$, then $x\in N$,
  • if $N\in N(x)$, $N\subseteq M$, then $M\in N(x)$,
  • if $N_1,N_2\in N(x)$, then $N_1\cap N_2\in N(x)$,
  • if $N\in N(x)$ there exists $M\subseteq N$, $x\in M$, such that $M\in N(y)$ for every $y\in M$,

I may define a topology on $X$ by saying that a subset $U\subseteq X$ is open if and only if, for every $x\in U$, there exists $N\in N(x)$ such that $N\subseteq U$. It is easy to see that these constructions are inverses to each other.

Giulio Bresciani
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