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Homeneighborhood system on a set

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# 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 systemof $X$ 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.

## Mathematics Subject Classification

54-00*no label found*

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