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 $x$.

Also see

 * Set is Open iff Neighborhood of all its Points