Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation

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

Let $T$ be such that there exists no continuous surjection from $T$ onto a discrete two-point space.

Then there exist no open sets $A, B \in \tau$ such that $A, B \ne \O$, $A \cup B = S$ and $A \cap B = \O$.

Proof
Let $T = \struct {S, \tau}$ be a topological space such that there exists no continuous surjection from $T$ onto a discrete two-point space.

Let $D = \struct {\set {0, 1}, \tau}$ be the discrete two-point space on $\left\{{0, 1}\right\}$.

Let $A$ and $B$ be disjoint open sets of $T$ such that $A \cup B = S$.

The aim is to show that one of them is empty.

Let us define the mapping $f: S \to \set {0, 1}$ by:


 * $\map f x = \begin{cases}

0 & : x \in A \\ 1 & : x \in B \end{cases}$

There are only four open sets in $\set {0, 1}$, namely:
 * $\O$
 * $\set 0$
 * $\set 1$
 * $\set {0, 1}$

We have that:


 * $f^{-1} \sqbrk \O = \O$


 * $f^{-1} \sqbrk {\set 0} = A$


 * $f^{-1} \sqbrk {\set 1} = B$


 * $f^{-1} \sqbrk {\set {0, 1} } = S$

where $f^{-1} \sqbrk X$ denotes the preimage of the set $X$.

But all of $\O, A, B, S$ are open sets of $T$.

So by definition $f$ is continuous.

Also, $f$ cannot be surjective.

It follows that $f$ must be constant.

So either $A$ or $B$ must be empty, and the other one must be $S$.

Hence the result.