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[parent] neighborhood system on a set (Definition)

In point-set topology, a neighborhood system is defined as the set of neighborhoods of some point in the topological space.

However, one can start out with the definition of a “abstract neighborhood system” $ \mathfrak{N}$ on an arbitrary set $ X$ and define a topology $ T$ on $ X$ based on this system $ \mathfrak{N}$ so that $ \mathfrak{N}$ is the neighborhood system of $ T$. This is done as follows:

Let $ X$ be a set and $ \mathfrak{N}$ be a subset of $ X\times P(X)$, where $ P(X)$ is the power set of $ X$. Then $ \mathfrak{N}$ is said to be a abstract neighborhood system of $ X$ if the following conditions are satisfied:
  1. if $ (x,U)\in \mathfrak{N}$, then $ x\in U$,
  2. for every $ x\in X$, there is a $ U\subseteq X$ such that $ (x,U)\in \mathfrak{N}$,
  3. if $ (x,U)\in \mathfrak{N}$ and $ U\subseteq V\subseteq X$, then $ (x,V)\in \mathfrak{N}$,
  4. if $ (x,U),(x,V)\in \mathfrak{N}$, then $ (x,U\cap V)\in \mathfrak{N}$,
  5. if $ (x,U)\in \mathfrak{N}$, then there is a $ V\subseteq X$ such that
    • $ (x,V)\in \mathfrak{N}$, and
    • $ (y,U)\in \mathfrak{N}$ for all $ y\in V$.
In addition, given this $ \mathfrak{N}$, define the abstract neighborhood system around $ x\in X$ to be the subset $ \mathfrak{N}_x$ of $ \mathfrak{N}$ consisting of all those elements whose first coordinate is $ x$. Evidently, $ \mathfrak{N}$ is the disjoint union of $ \mathfrak{N}_x$ for all $ x\in X$. Finally, let

$\displaystyle T$ $\displaystyle =$ $\displaystyle \lbrace U\subseteq X\mid$   for every $\displaystyle x\in U$, $\displaystyle (x,U)\in \mathfrak{N} \rbrace$  
  $\displaystyle =$ $\displaystyle \lbrace U\subseteq X\mid$   for every $\displaystyle x\in U$, there is a $\displaystyle V\subseteq U$, such that $\displaystyle (x,V)\in \mathfrak{N} \rbrace.$  

The two definitions are the same by condition 3. We assert that $ T$ defined above is a topology on $ X$. Furthermore, $ T_x:=\lbrace U\mid (x,U)\in \mathfrak{N}_x \rbrace$ is the set of neighborhoods of $ x$ under $ T$.
Proof. We first show that $ T$ is a topology. For every $ x\in X$, some $ U\subseteq X$, we have $ (x,U)\in \mathfrak{N}$ by condition 2. Hence $ (x,X)\in \mathfrak{N}$ by condition 3. So $ X\in T$. Also, $ \varnothing\in T$ is vacuously satisfied, for no $ x\in \varnothing$. If $ U,V\in T$, then $ U\cap V\in T$ by condition 4. Let $ \lbrace U_i\rbrace$ be a subset of $ T$ whose elements are indexed by $ I$ ($ i\in I$). Let $ U=\bigcup U_i$. Pick any $ x\in U$, then $ x\in U_i$ for some $ i\in I$. Since $ U_i\in T$, $ (x,U_i)\in \mathfrak{N}$. Since $ U_i\subseteq U$, $ (x,U)\in \mathfrak{N}$ by condition 3, so $ U\in T$.

Next, suppose $ \mathcal{N}$ is the set of neighborhoods of $ x$ under $ T$. We need to show $ \mathcal{N}=T_x$:

  1. ( $ \mathcal{N}\subseteq T_x$). If $ N\in\mathcal{N}$, then there is $ U\in T$ with $ x\in U\subseteq N$. But $ (x,U)\in \mathfrak{N}$, so by condition 3, $ (x,N)\in \mathfrak{N}$, or $ (x,N)\in \mathfrak{N}_x$, or $ N\in T_x$.
  2. ( $ T_x\subseteq \mathcal{N}$). Pick any $ U\in T_x$ and set $ W=\lbrace z\mid U\in T_z\rbrace$. Then $ x\in W\subseteq U$ by condition 1. We show $ W$ is open. This means we need to find, for each $ z\in W$, a $ V\subseteq W$ such that $ (z,V)\in \mathfrak{N}$. If $ z\in W$, then $ (z,U)\in \mathfrak{N}$. By condition 5, there is $ V\in \mathfrak{N}$ such that $ (z,V)\in \mathfrak{N}$, and for any $ y\in V$, $ (y,U)\in \mathfrak{N}$, or $ y\in U$ by condition 1. So $ y\in W$ by the definition of $ W$, or $ V\subseteq W$. Thus $ W$ is open and $ U\in \mathcal{N}$.
This completes the proof. By the way, $ W$ defined above is none other than the interior of $ U$: $ W=U^{\circ}$. $ \qedsymbol$

Remark. Conversely, if $ T$ is a topology on $ X$, we can define $ \mathfrak{N}_x$ to be the set consisting of $ (x,U)$ such that $ U$ is a neighborhood of $ x$. The the union $ \mathfrak{N}$ of $ \mathfrak{N}_x$ for each $ x\in X$ satisfies conditions $ 1$ through $ 5$ above:

  1. (condition 1): clear
  2. (condition 2): because $ (x,X)\in \mathfrak{N}$ for each $ x\in X$
  3. (condition 3): if $ U$ is a neighborhood of $ x$ and $ V$ a supserset of $ U$, then $ V$ is also a neighborhood of $ x$
  4. (condition 4): if $ U$ and $ V$ are neighborhoods of $ x$, there are open $ A,B$ with $ x\in A\subseteq U$ and $ x\in B\subseteq V$, so $ x\in A\cap B\subseteq U\cap V$, which means $ U\cap V$ is a neighborhood of $ x$
  5. (condition 5): if $ U$ is a neighborhood of $ x$, there is open $ A$ with $ x\in A\subseteq U$; clearly $ A$ is a neighborhood of $ x$ and any $ y\in A$ has $ U$ as neighborhood.

So the definition of a neighborhood system on an arbitrary set gives an alternative way of defining a topology on the set. There is a one-to-one correspondence between the set of topologies on a set and the set of abstract neighborhood systems on the set.



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Also defines:  abstract neighborhood system

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Cross-references: one-to-one correspondence, clear, union, conversely, interior, proof, completes, open, indexed by, vacuously, definitions, disjoint union, coordinate, addition, power set, subset, point, neighborhoods, neighborhood system, topology
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This is version 10 of neighborhood system on a set, born on 2007-02-12, modified 2007-02-19.
Object id is 8905, canonical name is NeighborhoodSystemOfASet.
Accessed 1175 times total.

Classification:
AMS MSC: 54-00 (General topology :: General reference works )
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