neighborhood system on a set

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

$\displaystyle T$ $\displaystyle=$ $\displaystyle\{U\subseteq X\mid\mbox{for every }x\in U\mbox{, }(x,U)\in% \mathfrak{N}\}$

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

The two definitions are the same by condition 3. We assert that $T$ defined above is a topology on $X$ . Furthermore, $T_{x}:=\{U\mid(x,U)\in\mathfrak{N}_{x}\}$ 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 $\{U_{i}\}$ 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 $W=\{z\mid U\in T_{z}\}$ . 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 $U$ : $W=U^{\circ}$ . ∎

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.

Title	neighborhood system on a set
Canonical name	NeighborhoodSystemOnASet
Date of creation	2013-03-22 16:41:34
Last modified on	2013-03-22 16:41:34
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	13
Author	CWoo (3771)
Entry type	Definition
Classification	msc 54-00
Defines	abstract neighborhood system

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