Equivalence of Definitions of Connected Topological Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Then the following definitions of connectedness are equivalent:

$(4) \implies (5)$: No Clopen Sets implies No Union of Separated Sets
=== $(5) \implies (6)$: No Union of Separated Sets implies No Surjection to Discrete Two-Point Space ===

No Surjection to Discrete Two-Point Space implies No Separation

 * $(6) \implies (1)$

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

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

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

Let us define the mapping $f: T \to \left\{{0, 1}\right\}$ by:


 * $f \left({x}\right) = \begin{cases}

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

There are only four open sets in $\left\{{0, 1}\right\}$, namely: $\varnothing$, $\left\{{0}\right\}$, $\left\{{1}\right\}$ and $\left\{{0, 1}\right\}$.

As:


 * $f^{-1} \left({\varnothing}\right) = \varnothing$


 * $f^{-1} \left({\left\{{0}\right\}}\right) = A$


 * $f^{-1} \left({\left\{{1}\right\}}\right) = B$


 * $f^{-1} \left({\left\{{0, 1}\right\}}\right) = T$

All of $\varnothing, A, B, T$ are open sets of $T$.

So by definition $f$ is continuous.

By $(6)$, $f$ cannot be surjective, so it must be constant.

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

Therefore, $\left\{{A, B}\right\}$ is not a separation of $T$.

Therefore $T$ admits no separation.

Also see

 * Condition on Connectedness by Clopen Sets for a separate proof that $(1)$ and $(4)$ are equivalent.