Set is Neighborhood of Subset iff Neighborhood of all Points of Subset
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $N \subseteq S$ be a subset of $T$.
Let $A \subseteq N$ be a subset of $T$.
Then:
- $N$ is a neighborhood of $A$ in $T$
- $N$ is a neighborhood of all points in $A$
Proof
Necessary Condition
Let $N$ be a neighborhood of $A$ in $T$.
By definition of neighborhood of set:
- $\exists U \in \tau : A \subseteq U \subseteq N$
Let $z \in A$.
By definition of subset:
- $z \in U$
From Set is Open iff Neighborhood of all its Points:
- $U$ is a neighborhood of $z$
From Superset of Neighborhood in Topological Space is Neighborhood:
- $N$ is a neighborhood of $z$
Since $z$ was arbitrary:
- $N$ is a neighborhood of all points in $A$
$\Box$
Sufficient Condition
Suppose that for all points of $z \in A$, $N$ is a neighborhood of $z$.
That is, for all $z \in A$ there exists an open set $U_z \subseteq N$ of $T$ such that $z \in U_z$.
Now by Union is Smallest Superset: Family of Sets:
- $\ds \bigcup_{z \mathop \in A} U_z \subseteq N$
as $\forall z \in A: U_z \subseteq N$.
If $z \in A$, then $z \in U_z$ by definition of $U_z$.
So:
- $\ds z \in \bigcup_{z \mathop \in A} U_z$
Thus we also have:
- $\ds A \subseteq \bigcup_{z \mathop \in A} U_z$
Let $U = \ds \bigcup_{z \mathop \in A} U_z$.
By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $U$ is open in $T$
Since $A \subseteq U \subseteq N$, then $N$ is neighborhood of $A$ in $T$ by definition.
$\blacksquare$