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 by hypothesis all of $\O, A, B, S$ are open sets of $T$.
So by definition $f$ is continuous.
Also by hypothesis, $f$ 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.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $6.2$: Connectedness: Proposition $6.2.3$