Neighborhood of All Points is Open

Theorem

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

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

Then:

$V$ is an open set of $T$

iff:

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

Proof

• Let $V$ be open in $T$.

Let $z \in V$.

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

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

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

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

• Now suppose that for all points of $z \in V$, $V$ is a neighborhood of $z$.

It can be seen that $V$ can be expressed as the union of all open sets contained in $V$.

Hence $V$ is open in $T$, by the axioms of a topological space.

$\blacksquare$

Sources

• Steven A. Gaal: Point Set Topology (1964)... (previous)... (next): $\S 1.1$: Lemma $1$
• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$