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.