# Equivalence of Definitions of Connected Topological Space

## Theorem

The following definitions of the concept of Connected Topological Space are equivalent:

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

### Definition 1

$T$ is connected if and only if it admits no separation.

### Definition 2

$T$ is connected if and only if it has no two disjoint nonempty closed sets whose union is $S$.

### Definition 3

$T$ is connected if and only if its only subsets whose boundary is empty are $S$ and $\O$.

### Definition 4

$T$ is connected if and only if its only clopen sets are $S$ and $\O$.

### Definition 5

$T$ is connected if and only if there are no two non-empty separated sets whose union is $S$.

### Definition 6

$T$ is connected if and only if there exists no continuous surjection from $T$ onto a discrete two-point space.

## Proof

### $(1) \iff (2)$: No Separation iff No Union of Closed Sets

From Biconditional Equivalent to Biconditional of Negations it follows that the statement can be expressed as:

$T$ admits a separation
there exist two closed sets of $T$ which form a (set) partition of $S$.

By definition, a separation of $T$ is a (set) partition of $S$ by $A, B$ which are open in $T$.

$A \mid B$ is a (set) partition of $S$ if and only if $\relcomp S A \mid \relcomp S B$ is a (set) partition of $S$.

Hence the result by definition of closed set.

$\Box$

### Definition by No Clopen Sets implies Definition by Separation

Let $T$ be connected by having no clopen sets.

Aiming for a contradiction, suppose $T$ admits a separation, $A \mid B$ say.

Then both $A$ and $B$ are clopen sets of $T$, neither of which is either $S$ or $\O$.

From this contradiction it follows that $T$ can admit no separation

$\Box$

### Definition by Separation implies Definition by No Clopen Sets

Let $T$ be connected by admitting no separation.

Suppose $\exists H \subseteq S$ which is clopen.

Then $\relcomp S H$ is also clopen.

Hence $H \mid \relcomp S H$ is a separation of $T$.

From this contradiction it follows that $T$ can have no proper subsets which are clopen.

$\Box$

### $(2) \implies (3)$: No Union of Closed Sets implies No Subsets with Empty Boundary

Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty.

Thus:

 $\ds \partial H$ $=$ $\ds \O$ by hypothesis $\ds \leadsto \ \$ $\ds H^- \cap \paren {S \setminus H}^-$ $=$ $\ds \O$ Boundary is Intersection of Closure with Closure of Complement

From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed sets of $T$.

$H^- \cup \paren {S \setminus H}^- = S$

Thus $H^-$ and $\paren {S \setminus H}^-$ are two disjoint closed sets of $T$ whose union is $S$.

Hence, by hypothesis, one of them must be empty.

Suppose $H$ is not empty.

$S \setminus H = \O$

Therefore $H = S$.

Thus the only subsets of $S$ whose boundary is empty are $S$ and $\O$.

$\Box$

### $(3) \implies (4)$: No Subsets with Empty Boundary implies No Clopen Sets

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

From Set is Clopen iff Boundary is Empty, $H$ has an empty boundary.

We have by hypothesis that $H = S$ or $H = \O$.

That is, the only clopen sets of $T$ are $S$ and $\O$.

$\Box$

### $(4) \implies (5)$: No Clopen Sets implies No Union of Separated Sets

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

By definition of separated sets:

$A \cap B^- = \O$

Then:

 $\ds S$ $=$ $\ds A \cup B$ $\ds$ $\subseteq$ $\ds A \cup B^-$ Set is Subset of its Topological Closure $\ds$ $\subseteq$ $\ds S$ by definition of $S$

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

From Topological Closure is Closed, $B^-$ is closed in $T$.

Thus $A$ is open in $T$.

Also by definition of separated sets:

$A^- \cap B = \O$

Hence, by the same reasoning, $B$ must also be open.

But:

$A \cap B \subseteq A \cap B^- = \O$

and $A \cup B = S$, by assumption.

So:

$A = S \setminus B$ and $B = S \setminus A$

and we conclude that both $A$ and $B$ are clopen.

Therefore, by hypothesis, one of them must be $S$ and the other must be $\O$.

That is, there are no two non-empty separated sets of $T$ whose union is $S$.

$\Box$

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

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

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

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

By definition of continuous mapping:

$\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are open sets of $T$.

From the definition of a mapping:

$\map {f^{-1} } 0 \cup \map {f^{-1} } 1 = S$

and

$\map {f^{-1} } 0 \cap \map {f^{-1} } 1 = \O$

Then:

$\map {f^{-1} } 0 = S \setminus \map {f^{-1} } 1$

and:

$\map {f^{-1} } 1 = T \setminus \map {f^{-1} } 0$

are clopen.

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

It follows from the definition that $\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ 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.

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

### $(6) \implies (1)$: No Continuous Surjection to Discrete Two-Point Space implies No Separation

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.

$\blacksquare$