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

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Theorem

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

Let $T$ be such that there are no two non-empty separated sets whose union is $S$.


Then there exists no continuous surjection from $T$ onto a discrete two-point space.


Proof

Let $T = \left({S, \tau}\right)$ be a topological space such that there are no two non-empty separated sets whose union is $S$.

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

Aiming for a contradiction, suppose $f: T \to \left\{{0, 1}\right\}$ is a continuous surjection.

By definition of continuous mapping:

$f^{-1} \left({0}\right)$ and $f^{-1} \left({1}\right)$ are open sets of $T$.


From the definition of a mapping:

$f^{-1} \left({0}\right) \cup f^{-1} \left({1}\right) = S$

and

$f^{-1} \left({0}\right) \cap f^{-1} \left({1}\right) = \varnothing$


Then:

$f^{-1} \left({0}\right) = S \setminus f^{-1} \left({1}\right)$

and:

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

are clopen.

From Closed Set equals its Closure they are their respective closures.


It follows from the definition that $f^{-1} \left({0}\right)$ and $f^{-1} \left({1}\right)$ are separated subsets of $T$ whose union is $S$.

Hence, by hypothesis, one of them must be empty, and the other one must be $S$.

Therefore $f$ is constant, and so is not a surjection.

This contradicts the original hypothesis.


That is, there exists no continuous surjection from $T$ onto a discrete two-point space.

$\blacksquare$