08. Connectedness and Path-Connectedness

8.1 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 $X$

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Connected

Let $X$ be a metric space then

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

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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 $X$

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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

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

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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

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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

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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

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Preservation of Connectedness Under Homeomorphisms

Property of Connectedness is preserved under Homeomorphisms

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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

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11 - Connectedness of Intervals

Connectedness of Intervals

Any interval $[x,y]$ in $\mathbb{R}$ is connected

Note that IVP is a consequence of this theorem and Connected image of a connected set

Proof

Using Connectedness via open separations
Suppose $[x,y]\subseteq U\cup V$ where $U$ and $V$ are open subsets in $\mathbb{R}$ with $[x,y]\cap U\cap V=\varnothing$
And $[x,y]⊈U$ and $[x,y]⊈V$ thus

$$[x,y]\cap U\text{ and }[x,y]\cap V\text{ are both non-empty}$$

Without loss of generality assume $x\in [x,y]\cap X$ and $y\in [x,y]\cap V$

Why $y\in [x,y]\cap V$

As $[x,y]\cap V\ne \varnothing$ there exists $y^{\prime }\in [x,y]$ with $y^{\prime }\in V$
Note $y^{\prime }\ne x\in U$

Replacing $[x,y]$ by possibly smaller interval $[x,y^{\prime }]$ then

$$[x,y^{\prime }]\subseteq U\cup V\text{ with }[x,y^{\prime }]\cap U\cap V=\varnothing$$

Moreover

$$[x,y^{\prime }]⊈U\text{ and }[x,y^{\prime }]⊈V\text{ as }x\in U\text{ and }{y}^{\prime }\in V$$

This leads to a contradiction to show assumption hence $y\in [x,y]\cap V$

Note that as $[x,y]\cap U\cap V=\varnothing$ then $y\notin U$ and $x\notin V$

Define $S=\{z\in [x,y]:z\in U\}$ then $S$ is non-empty and bounded so

$$c=\sup (S)\text{ exists}$$

Note that $c\in [x,y]$ and since $[x,y]\subseteq U\cup V$ then either $c\in U$ or $c\in V$

Assume $c\in U$ so as $y\notin U$ then $c\ne y$
As $U$ is open then there is some interval $[c,c+ϵ)$ contained in $U$ and also in $[x,y]$
However $[c,c+ϵ)\subseteq S$ but contradicts that $c=\sup (S)$

Similarly assume if $c\in V$ then $c\ne x$
As $V$ is open there is some interval $(c-ϵ,c]$ contained in $V$ and in $[x,y]$
In particular as $\left[c-\frac{ϵ}{2},c\right]$ is disjoint from $S$ this contradicts $c=\sup S$

Hence as we have the two contradictions then neither $[x,y]\subseteq U$ or $[x,y]\subseteq V$ hence

$$[x,y]\text{ is connected}$$

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8.2 Path-Connectedness

Path Connectedness

Let $X$ be a metric space then

$X$ is path-connected if
For $a,b\in X$ there is a continuous map $\gamma :[0,1]\to X$ with

$$\gamma (0)=a\text{ and }\gamma (1)=b$$

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Path

Let $X$ be a metric space then

Continuous Map $\gamma :[0,1]\to X$ is called a path

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Concatenation of Paths

Let $X$ be a metric space with two paths ${\gamma }_1,{\gamma }_2$ such that

$${\gamma }_1(1)={\gamma }_2(0)$$

then concatenation $y_1\star y_2$ of the two paths is defined by

$${\gamma }_1\star {\gamma }_2(t)=\{\begin{array}{ll}{\gamma }_1(2t)& \text{if }0\le t\le \frac{1}{2}\\ & \\ {\gamma }_2(2t-1)& \text{if }\frac{1}{2}\le t\le 1\end{array}$$

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Opposite Path

Let $X$ be a metric space with path $\gamma$ then

Opposite path ${\gamma }^{-1}$ is defined by

$${\gamma }^{-1}(t)=\gamma (1-t)$$

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Equivalence Relation between Elements of a Metric Space

Let $X$ be a metric space

Define relation $\sim$ on $X$ as follows

$$a\sim b⟺\text{exists path }\gamma :[0,1]\to X\text{ with }\gamma (0)=a\text{ and }\gamma (1)=b$$

then

$$\sim \text{ is an equivalence relation}$$

Proof

$a\sim a$ by $\gamma (t)=a$ which takes constant value $a$
$a\sim b$ implies $b\sim a$ by taking path $\gamma$ from $a$ to $b$ and considering opposite path $g^{-1}$
$a\sim b\text{ and }b\sim c$ implies $a\sim c$ use concatenation of the two paths

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Path-Components

Let $X$ be a metric space then

Equivalence classes in which ~ partitions $X$ is called path-components of $X$

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8.3 Connectedness and Path-Connectedness

12 - Path Connected and Connected

Path Connected and Connected

A path-connected metric space is connected

Proof

Suppose that $X$ is path-connected

Let $f:X\to \{0,1\}$
Let $a,b\in X$ so as $X$ is path connected there exists path $\gamma :[0,1]\to X$ such that

$$\gamma (0)=a,\gamma (1)=b$$

Consider composition $f\circ \gamma$ which is a continuous function from $[0,1]$ to $\{0,1\}$
Hence as $[0,1]$ is connected by Connectedness of Intervals then $f\circ \gamma$ is constant
Thus

$$f(a)=(f\circ \gamma )(0)=(f\circ \gamma )(1)=f(b)$$

As $a$ and $b$ are arbitrary then $f$ is constant
Hence by Equivalent statements for connected metric spaces then $X$ is connected

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13 - Path-Connectedness of a Connected Open Subset of a Normed Space

Path-Connectedness of a Connected Open Subset of a Normed Space

A connected open subset of a normed space is path-connected

Proof

Let $X$ be the connected open set

Suppose $P$ is a path-component of $X$
Let $a\in P$

Since $X$ is open then there exists ball $B(a,ϵ)$ contained in $X$
Let $b\in B(a,ϵ)$

So define explicit path $\gamma$ between $a$ and $b$ by

$$\gamma (t)=(1-t)a+tb$$

This is continuous and the image is contained in $B(a,ϵ)$ since

$$||\gamma (t)-a||=t||a-b||\le ||a-b||=d(a,b)<ϵ$$

for all $t\in [0,1]$

Hence $b\in P$ as it lies in the same path-component
So we have the path-components partition $X$ hence if there is more than one then
$X$ can be written as a disjoint union of non-empty open sets contrary to assumption

Hence there is only one path-component which is $X$ so $X$ is path-connected

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14 - Connected but not Path-Connected Set in ℝ²

Connected but not Path-Connected Set in ${\mathbb{R}}^2$

There is a connected subset of ${\mathbb{R}}^2$ which is not path-connected

Proof - Counter-Example

Consider Topologist’s Sine Curve given by

$$\{(0,y):-1\le y\le 1\}\cup \left\{\left(x,\sin \left(\frac{1}{x}\right):x\in (0,1]\right)\right\}$$

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