Space is Neighborhood of all its Points

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $x \in S$.

Then $S$ is a neighborhood of $x$.

Proof

By the definition of the topology $\tau$, $S$ is an open set.

From Set is Open iff Neighborhood of all its Points, $S$ is a neighborhood of $x$.

$\blacksquare$

Also see

• Set is Open iff Neighborhood of all its Points