If $G\subset \mathbb{R}^p$ is open then $G$ is connected iff G is polygonally path-connected.

I am having trouble understanding the proof of this theorem.


Assume $G$ is connected and choose any $x\in G$. Let$$\begin{cases} G_1=\{y\in G:\text{y can be joined to x by a polygonal path contained in G}\} \\\\ G_2 =\{z\in G:\text{z cannot be joined to x by a polygonal path contained in G}\} \end{cases}$$ $(1)$ Note that $G_1\cup G_2=G\ \text{and}\ G_1\cap G_2 =\emptyset$.

Choose $y\in G_1$ since $G$ is open. Since $G$ is open, $\exists\ r>0$ such that $B_r(y)\subset G$. $(2)$ Choose $z\in B_r(y)$; we must show $z\in G_1$. Let $P$ be the polygonal curve joining $x$ to $y$.

(3) Note $y$ is joined to $z$ by a line segment $L$ lying in $G$, hence $P\cup L$ is a polygonal curve joining $x$ to $z$. Hence $z \in G_1$ so $G_1$ is open. $(4)$ Claim $G_2$ is open, thus $(G_1,G_2)$ is a disconnection of $G$ if $G_1,G_2\ne \emptyset$. Hence one of $G_1,G_2$ must be empty which is $G_2 =\emptyset$ since $x\in G_1$.

For $(1)$, why is $G_1\cap G_2 =\emptyset$, aren't we suppose to show that it is connected thus it can't be disjoint? For $(2)$, we are able to let $z \in B_r(y)$ because of the fact it is open, correct? For $(3)$ why is there a line segment joining $x$ to $z$; in the conditions it says that $z$ cannot be joined to $x$? And $(4)$, why is it a disconnection if $G_1,G_2\ne \emptyset$?


(1) By definition, $G_2=G\setminus G_1,$ so they are disjoint.

(2) No. We can choose $z\in B_r(y)$ because it is non-empty. We are trying to show that $G_1$ is open by showing that for each $y\in G_1$ there is an open ball around $y$ contained in $G_1$. Knowing that $G$ is open allows us to pick a candidate ball.

(3) We can only conclude that $z$ can't be connected to $x$ if $z\in G_2$. Simply calling a point "$z$" isn't enough to conclude that it lies in $G_2$. If it's confusing you, define $$G_2=\{w\in G:w\text{ cannot be joined to }x\text{ by a polygonal path contained in }G\}$$ instead, and proceed with the rest of the proof in the same way.

(4) A disconnection of $G$ is two non-empty disjoint open sets whose union is $G$. The proof showed that they were both open, and were disjoint, and that their union was $G$. Hence, they are a disconnection if they are both non-empty.

  • $\begingroup$ I understand $1$ and $4$ now, but I still dont understand $2$ and $3$, more so $2$. $\endgroup$ – Q.matin Mar 25 '13 at 5:23
  • $\begingroup$ For (3), note that we can join $z$ to $y$ with a line segment $L$ lying in $B_r(y),$ since $z\in B_r(y).$ As $B_r(y)\subseteq G$, then $L\subseteq G$, and since $P\subseteq G$ (by choice of $P$, which choice is possible since $y\in G_1$), then.... $\endgroup$ – Cameron Buie Mar 25 '13 at 5:35
  • $\begingroup$ As for (2), note that $B_r(y)\subseteq G$ turned out to be enough to ensure that we could connect any point in $B_r(y)$ to $x$ with a polygonal path. Think on it, and let me know if you still have questions. $\endgroup$ – Cameron Buie Mar 25 '13 at 5:36
  • $\begingroup$ I understand $3$ now, but $2$ is still confusing. Why are we allowed to pick $z\in B_r(y)$? Is it because since $z\in G$ and $B_r(y)\subset G$? $\endgroup$ – Q.matin Mar 25 '13 at 5:45
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    $\begingroup$ No problem! Glad to help. $\endgroup$ – Cameron Buie Mar 25 '13 at 5:55

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