19 - Connectedness

Disconnected

Let $X$ be a metric space then

$X$ is disconnected if it can be written as the disjoint union of two non-empty open sets

Note that if $X$ is written as a disjoint of two non-empty sets $U$ and $V$ then $U,V$ disconnect $X$

Connected

Let $X$ be a metric space then

$X$ is connected if $X$ is not disconnected

Equivalent Statements for Connected Metric Spaces lemma

Let $X$ be a metric space then the following statements are equivalent

1. $X$ is connected
2. If $f:X\to \{0,1\}$ is a continuous function then

$$f\text{ is constant}$$

3. The only subsets of $X$ which both open and closed are $X$ and $\varnothing$

Proof

$(1)⟹(2)$
Let $X$ be connected and let $f:X\to \{0,1\}$ be a continuous function
Singleton sets $\{0\}$ and $\{1\}$ are both open in $\{0,1\}$ so

$$f^{-1}(0)\text{ and }{f}^{-1}(1)\text{ are open subsets of X }$$

They are disjoint and their union is $X$ hence one of them must be empty so $f$ is constant

$(2)⟹(3)$
Suppose $A\subseteq X$ is open and closed then $A^c$ is open and closed
So function $f:X\to \{0,1\}$ defined by $f(x)=\{\begin{array}{ll}1& x\in A\\ 0& x\in A^c\end{array}$
By $(ii)$ then $f$ must be constant
If $f(x)=1$ then $A=X$
If $f(x)=0$ then $A=\varnothing$

$(3)⟹(1)$
Suppose that $X=U\cup V$ with $U,V$ open and disjoint then
$U^c=V$ is open so $U$ is closed
As $U$ is both open and closed then it is either $X$ or $\varnothing$ and similarly for $V$
Hence there is no way to disconnect $X$

Connectedness via Open Separations lemma

Let $X$ be a metric space and let $Y\subseteq X$ be a subset (induced metric space from $X$) then

$Y$ is connected if and only if for open subsets $U,V$ of $X$ and $U\cap V\cap Y=\varnothing$ then

$$Y\subseteq U\cup V⟹Y\subseteq U\text{ or }Y\subseteq V$$

Proof

Open sets in $Y$ are precisely sets of form $U\cap Y$ where $U$ is open in $X$
by Property of closed and open subsets between subspaces

Take pair $U\cap Y$ and $V\cap Y$ of such open sets so they disconnect $Y$ if and only if

1. They are disjoint so

$$(U\cap Y)\cap (V\cap Y)=U\cap V\cap Y=\varnothing$$

2. They cover $Y$ so

$$Y\subseteq U\cup V$$

3. Neither is empty

$$U\ne \varnothing \text{ and }V\ne \varnothing$$

Hence $Y$ is connected if and only if $(1)$ and $(2)$ imply that

$$U\cap Y=\varnothing \text{ or }V\cap Y=\varnothing$$

Hence either

$$Y\subseteq V\text{ or }Y\subseteq U$$

Sunflower Lemma lemma

Let $X$ be a metric space
Let $\{A_i:i\in I\}$ be a collection of connected subsets of $X$ such that

$$\bigcap _{i\in I}{A}_i\ne \varnothing$$

then

$$\bigcup _{i\in I}{A}_i\text{ is connected}$$

Proof

Using Equivalent statements for connected metric spaces (2)
Suppose $f:{\bigcup }_{i\in I}{A}_i\to \{0,1\}$ is continuous

Pick $x_0\in {\bigcap }_{i\in I}{A}_i$ then if $x\in {\bigcup }_{i\in I}{A}_i$ there exists $i\in I$ such that $x\in A_i$
By restriction of $f$ to $A_i$ being constant since $A_i$ is connected then

$$f(x)=f(x_0)\text{ as }x,x_0\in A_i$$

As $x$ was arbitrary then $f$ is constant

Connectedness and Closures lemma

Let $X$ be a metric space

If $A\subseteq X$ is connected then if $B$ is such that

$$A\subseteq B\subseteq \overline{A}$$

Then

$$B\text{ is connected}$$

Proof

Using Connectedness via open separations so
Suppose that $B\subseteq U\cup V$ where $U$ and $V$ are open in $X$ and $U\cap V\cap B=\varnothing$
So as $A\subseteq B$ then similarly $A\subseteq U\cup V$ and $A\cap U\cap V=\varnothing$

As $A$ is connected then either

$$A\subseteq U\text{ or }A\subseteq V$$

Without loss of generality then $A\subseteq U$ and since $A\cap U\cap V=\varnothing$ then

$$A\subseteq V^c$$

As $V$ is closed then by taking closures

$$\overline{A}\subseteq \overline{V^c}=V^c$$

Hence $B\subseteq V^c$ so since $B\subseteq U\cup V$ then

$$B\subseteq U$$

Hence $B$ is connected

Connected Image of a Connected Set lemma

Let $X$ be a connected metric space
Let $f:X\to Y$ be continuous then

$$f(X)\text{ is connected}$$

Proof

Suppose $f$ is surjective (otherwise replace $Y$ by $f(X)$)
Suppose $U$ and $V$ are disjoint open subsets of $Y$ with $U\cup V=Y$ then

$$f^{-1}(U)\text{ and }{f}^{-1}(V)\text{ are disjoint open subsets of }X$$

with

$$f^{-1}(U)\cup f^{-1}(V)=X$$

Since $X$ is connected then one of them must be empty
Without loss of generality then $f^{-1}(U)$ is empty hence $U$ is empty
Hence $Y$ cannot be disconnected

Preservation of Connectedness Under Homeomorphisms

Property of Connectedness is preserved under Homeomorphisms

Connected Components

The connected components of a metric space partition the space
So

$$\text{Space is connected}⟺\text{it has a unique connected component}$$

where for each $x\in X$ the connected component of $X$ containing $x$ is the

$$\text{unique maximal connected subset of }X\text{ containing }x$$

Proof

Let $X$ be the space and for $x\in X$ let $\Gamma (x)$ be the connected component containing $x$
Suppose $\Gamma (x)$ and $\Gamma (y)$ are not disjoint so exist $a\in \Gamma (x)\cap \Gamma (y)$

By Sunflower Lemma then $\Gamma (x)\cup \Gamma (y)$ is connected
By definition of connected component then $\Gamma (x)\cup \Gamma (y)\subseteq \Gamma (x)$ so

$$\Gamma (y)\subseteq \Gamma (x)$$

and similarly

$$\Gamma (x)\subseteq \Gamma (y)$$

Hence

$$\Gamma (x)=\Gamma (y)$$

Hence the connected components partition the space