LEC ANA2 14

Path connectedness

Definition 337.1(Path connectedness).

Let $X$ be a metric space. Let $x,y\in X$. A path from $x$ to $y$ is a continuous function $\gamma :[0,1]\to X$ such that $\gamma (0)=x$, $\gamma (1)=y$. If there exists a path between $x$ and $y$, the points are said to be path connected. $S\subseteq X$ is called path connected if every pair of points $x,y\in S$ are path connected.

The relation induced by path connectedness is an equivalence relation.

Proposition 337.2(Proposition).

Path connected $⟹$ connected.

Proof.

Suppose $X$ is path connected. FTSOC, suppose $X=A\bigsqcup B$, where $A$ and $B$ are nonempty and open in $X$. Let $a\in A$ and $b\in B$. Let $\gamma :[0,1]\to X$ be a path from $a$ to $b$. $\gamma ([0,1])=(\gamma ([0,1])\cap A)\bigsqcup (\gamma ([0,1])\cap B)$ is a separation of $\gamma ([0,1])$, which is supposed to be a connected set by Thm 144.4.□

Example 337.3(Example).

Let $S\subseteq {\mathbb{R}}^n$ be countable. Then, ${\mathbb{R}}^n\setminus S$ is path connected.
This follows from the fact that there are uncountably many disjoint paths between any two fixed points in ${\mathbb{R}}^n$; for instance, there are uncountably many circles passing through any two given points.

Example 337.4(Example).

Open balls in any NLS are path connected. Indeed, if $x,y\in B_r(x_0)$, then for all $0⩽t⩽1$,

$$\begin{array}{rl}\Vert (tx+(1-t)y)-x_0\Vert & ⩽\Vert t(x-x_0)\Vert +\Vert (1-t)(y-x_0)\Vert \\ & ⩽tr+(1-t)r\\ & =r,\end{array}$$

so $tx+(1-t)y\in B_r(x_0)$. It follows that $\gamma (t)=tx+(1-t)y$ is a path between $x$ and $y$.

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Proposition 337.5(Proposition).

Let $X$ be a NLS. If $U\subseteq X$ is open and connected, it is path connected.

Proof.

Fix$x_0\in U$. Let $E=\{x\in U:x,x_0\text{ are path connected}\}$. It suffices to prove $E$ is both open and closed in $U$.

Let $x\in E$. Since $U$ is open, there exists an open ball $B$ centered at $x$ contained in $U$; it follows from Exm 4 that $B\subseteq E$. Therefore, $E$ is open. The same argument shows $E^c$ is open.□

Example 337.6(Example).

Clearly, if $E$ is connected, any $F$ such that $E\subseteq F\subseteq \overline{E}$ is connected - just apply Prp 329.1. Here’s an example where $E$ is path connected, $\overline{E}$ is path connected, and $F$ is connected (but not path connected!):

$$\begin{array}{rl}& K:=\bigcup _{i=1}^{\infty }\left(\left\{1/i\right\}\times [0,1]\right),\\ & E:=K\cup ([0,1]\times \{0\}),\\ & \overline{E}=E\cup (\{0\}\times [0,1]),\\ & F:=E\cup \{(0,1)\}.\end{array}$$

Clearly, $E$ and $\overline{E}$ are path connected. Suppose $\gamma :[0,1]\to F$ is a path from $(0,1)$ to $(1,0)$. Let $t_0=\sup {\gamma }^{-1}(\{(0,1)\})$. We claim that there exists $ϵ>0$ such that $\gamma ((t_0,t_0+ϵ))\subseteq K$. If such an $ϵ$ did not exist, we can construct a sequence $\{t_n\}\to t_0$ such that $t_n>t_0$ and ${\pi }_y(\gamma (t_n))\to 0$, yielding a contradiction.

Since $(t_0,t_0+ϵ)$ is connected, we must have $\gamma ((t_0,t_0+ϵ))\subseteq \{1/n_0\}\times [0,1]$ for some $n_0\in \mathbb{N}$. For any $\{t_n\}\to t_0$ such that $t_n>t_0$, we have ${\pi }_x(\gamma (t_n))=1/n_0\ne 0$, a contradiction. Thus, there does not exist a path from $(0,1)$ to $(1,0)$ in $F$.

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