Set is Neighborhood of Subset iff Neighborhood of all Points of Subset

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 $A$.
 * $N$ is a neighborhood of all points in $A$.

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 $z$

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 :
 * $U$ is open in $T$

Since $A \subseteq U \subseteq N$, then $N$ is neighborhood of $A$ in $T$ by definition.