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 $T$

if and only if:

$N$ 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.

$\blacksquare$