Equivalence of Definitions of Connected Topological Space

Theorem
Let $T$ be a (nonempty) topological space.

Then the following definitions of connectedness:


 * $(1): \quad T$ is connected iff it admits no partition.


 * $(2): \quad T$ is connected iff it has no two disjoint nonempty closed sets whose union is $T$.


 * $(3): \quad T$ is connected iff its only subsets whose boundary is empty are $T$ and $\varnothing$.


 * $(4): \quad T$ is connected iff its only clopen sets are $T$ and $\varnothing$.


 * $(5): \quad T$ is connected iff there are no two nonempty separated sets whose union is $T$.


 * $(6): \quad T$ is connected iff there does not exist any continuous surjection from $T$ onto a discrete two-point space

are equivalent.

Proof

 * $(1) \implies (2)$

Suppose that $A, B \subseteq T$ are closed sets of $T$ such that $A \cap B = \varnothing$ and $A \cup B = T$.

Then $A = T \setminus B$ and $B = T \setminus A$ are both open sets whose union is $T$.

If $A$ and $B$ were nonempty, $A \mid B$ would be a partition of $T$, contradicting $(1)$.

Hence, $A$ or $B$ must be empty, and $(2)$ is valid.


 * $(2) \implies (3)$

Let $X \subseteq T$ be a nonempty subset whose boundary is empty, that is:


 * $\partial X = X^- \cap \left({T \setminus X}\right)^- = \varnothing$

But then, $X^-$ and $\left({T \setminus X}\right)^-$ are two closed sets of $T$ whose union is $T$.

Hence, one of them must be empty, by $(2)$.

By assumption, $X$ is not empty, and then, $T \setminus X = \varnothing$.

Therefore $X = T$, and it follows that $(3)$ is valid.


 * $(3) \implies (4)$

Let $S \subseteq T$ be a clopen set of $T$.

Then by definition $T \setminus S$ is also clopen.

Hence, both $S$ and $T \setminus S$ are their closures, and:


 * $\partial S = S^- \cap \left({T \setminus S}\right)^- = S \cap \left({T \setminus S}\right) = \varnothing$

Then $S$ has an empty boundary, and by $(3)$, $S = T$ or $S = \varnothing$.

Therefore, $(4)$ is valid.


 * $(4) \implies (5)$

Suppose $A$ and $B$ are separated subsets of $T$ such that $A \cup B = T$.

Then $A \cap B^- = \varnothing$ and $T = A \cup B \subseteq A \cup B^- \subseteq T$.

Hence $A = T \setminus B^-$.

As $B^-$ is closed, A is open.

By the same reasoning, $B$ must also be open.

But $A \cap B \subseteq A \cap B^- = \varnothing$, and $A \cup B = T$, by assumption.

So $A = T \setminus B$ and $B = T \setminus A$, and we conclude that both $A$ and $B$ are clopen.

Therefore, by $(4)$, one of them must be $T$ and the other must be $\varnothing$, concluding $(5)$.


 * $(5) \implies (6)$

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

Suppose $f: T \rightarrow \left\{{0, 1}\right\}$ is a continuous mapping.

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

From the definition of a mapping:
 * $f^{-1} \left({0}\right) \cup f^{-1} \left({1}\right) = T$

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

Then, $f^{-1} \left({0}\right) = T \setminus f^{-1} \left({1}\right)$ and $f^{-1} \left({1}\right) = T \setminus f^{-1} \left({0}\right)$ are clopen, and 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 $T$.

Hence, by $(5)$, one of them must be empty, and the other one must be $T$.

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

So $(6)$ is valid.


 * $(6) \implies (1)$

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 partition of $T$.

Therefore $T$ admits no partition.