Set is Open iff Neighborhood of all its Points

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $V \subseteq S$ be a subset of $T$.


Then:

$V$ is an open set of $T$

if and only if:

$V$ is a neighborhood of all the points in $V$.


Proof

Necessary Condition

Let $V$ be open in $T$.

Let $z \in V$.

By definition, a neighborhood of $z$ is any subset of $S$ containing an open set which itself contains $z$.

But $V$ is itself an open set which itself contains $z$.

Hence by Set is Subset of Itself, $V$ is a subset of $S$ which contains an open set which itself contains $z$.

So for all points of $z \in V$, $V$ is a neighborhood of $z$.

$\Box$


Sufficient Condition

Suppose that for all points of $z \in V$, $V$ is a neighborhood of $z$.

That is, for all $z \in V$ there exists an open set $T_z \subseteq V$ of $T$ such that $z \in T_z$.


Now by Union is Smallest Superset: Family of Sets:

$\displaystyle \bigcup_{z \mathop \in V} T_z \subseteq V$

as $\forall z \in V: T_z \subseteq V$.


If $z \in V$, then $z \in T_z$ by definition of $T_z$.

So:

$\displaystyle z \in \bigcup_{z \mathop \in V} T_z$

Thus we also have:

$\displaystyle V \subseteq \bigcup_{z \mathop \in V} T_z$

Hence by definition of set equality:

$\displaystyle V = \bigcup_{z \mathop \in V} T_z$


Thus $V$ can be expressed as a union of open sets.

Hence $V$ is open in $T$, by axiom $(O1)$ of a topological space.

$\blacksquare$


Also see


Sources