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.

  • 3
    $\begingroup$ Looks good to me. There are some wording issues ( you say "For $X$ to be true, then $Y$." Then you prove $Y$ and conclude $X$. This is not a logically correct. However, what you meant was " $X$ is equivalent to $Y$" ), but the mathematics is good. $\endgroup$ Apr 4, 2015 at 6:01
  • $\begingroup$ When you feel comfortable enough with these examples, check this out: math.stackexchange.com/questions/843173/… $\endgroup$ Apr 4, 2015 at 7:35
  • 2
    $\begingroup$ Why A is non empty? You said x is in A but here x is a arbitrary element.not a fixed element.I think it should be 'a' . a is in A and hence A is non empty. $\endgroup$
    – LAMDA
    May 15, 2020 at 3:02
  • $\begingroup$ @LAMDA right?’ That’s what I thought too , we have $a \in A$ not $x$. $\endgroup$
    – anonymous
    Aug 27, 2022 at 3:55

1 Answer 1


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.


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