The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.
I am to prove the following:
Prove that every connected open $U\subset \mathbb{R}^2$ is path connected.
This is what I understand:
- A space $X$ is connected provided that it cannot be written as the disjoint union of specifically two open sets.
- A space $X$ is path connected if given $a,b\in X$, there is a continuous function $f:[0,1]\rightarrow X$ such that $f(0)=a$ and $f(1)=b$
If I am not mistaken, since $U$ is connected then the only sets which are clopen are $\emptyset$ and $U$. Therefore, given any $a\in U$, I would need to show a set of points, say $A\subseteq U$, which can be joined to $a$ by a path in $U$ is clopen.
Rough Proof:
Suppose $U$ is a connected open subset of $\mathbb{R}^2$. Let $a\in U$ and $A\subseteq U$ be a set of points which can join $a$. We define $C=U \setminus A$; therefore, $A\cap C=\emptyset$ by set difference. We seek to prove $A \neq \emptyset$, $A$ is open, and $A$ is closed.
Let $x\in A$ (hence not empty); suppose $\exists \epsilon >0 : B(x,\epsilon)\subseteq U$. Since open balls are convex, it is path connected. Thus for any point $y$ in $B(x)$, there is a path from $x$ to $y$. Since $A$ is a set of points in $U$ that can join $a$, then there exist a path between $x$ and $a$. Since there is a path from $y$ to $x$ and $x$ to $a$; then, there is a path from $y$ to $a$ implying $y$ is in $A$. Since $y\in B(x,\epsilon)$, we conclude $B(x) \subseteq A$, hence $A$ is open.
By definition, for $A$ to be closed, then $U\setminus A = C$ must be open. Using a similar argument above, if $x\in C$, then $\exists \epsilon>0$: $B(x,\epsilon)\subseteq U$. If any point $y\in B(x)$ joins $a$, so can $x$. Hence $C$ is open, therefore $A$ is closed.
By hypothesis, $U$ is connected. Since $A$ is a nonempty subset of $U$, then if follows that $C=\emptyset$ and $A=U$; therefore $U$ is path connected.
Am I on the right track? Is there anything I need to revise or make clear in my proof?
I sincerely thank you for taking the time to read this question and my attempt at proving it. I greatly appreciate any assistance you may provide.
Best Answer
The above proof is almost correct. The set $C$ should be the set of points that cannot be joined with $a$ by a continuous path within $U$. $C$ is open since every $y\in B_r(x)$ cannot be joined with $a$ for $r>0$ such that $B_r(x)\subset U$. Otherwise, we could join $a$ with $y$ and $y$ with $x$ by linear path so $x$ could be joined with $a$. That is a contradiction.