Non-Trivial Discrete Space is not Connected

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space where $\tau$ is the discrete topology on $S$.

Let $T$ be non-trivial.

Then $T$ is not connected.

Thus $T$ is not:
 * arc-connected


 * path-connected


 * irreducible


 * ultraconnected

Proof
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete space.

Let $a \in S$.

Let $A = \left\{{a}\right\}$ and $B = \complement_S \left({\left\{{a}\right\}}\right)$, where $\complement_S \left({\left\{{a}\right\}}\right)$ is the complement of $A$ in $S$.

As $T$ is not trivial, $B \ne \varnothing$.

Then from Set in Discrete Topology is Clopen, $A$ and $B$ are both open.

So $A \mid B$ is a separation of $S$.

It follows, by definition, that $T$ is not connected.

From Sequence of Implications of Connectedness Properties, it follows that $T$ has none of those other connectedness properties either.