3
$\begingroup$

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.

Proof:

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 $G_1,G_2\ne \emptyset$?

$\endgroup$
3
$\begingroup$

(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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ No problem! Glad to help. $\endgroup$ – Cameron Buie Mar 25 '13 at 5:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.