# Show that an open connected subset of R^2 is path-connected

1. Jul 29, 2013

Hi!

I have a question regarding my solution to a problem in topology.

Problem: Show that if U is an open connected subset of ℝ2, then U is also path-connected.

Hint: Show that given any x0 in U, show that the set of points that can be joined to x0 by a path in U is both open and closed.

First of all, I know that since U is connected, the only sets which are both open and closed are ∅ and U itself. Therefore, if I can show that the set of points which can be joined to x0 by a path is indeed open and closed, and non-empty, it is U. But I am stuck, and think I found an alternative solution to this problem, not using the hint, and was wondering if it is correct or not.

My idea is this. Since U is an open subset of ℝ2, U can be written as a (possibly infinite) union of open balls Bn(x,ε), centered at x and with radius ε, positive, where x lies in U. And U is connected, which implies that given any ball, there is another ball with nonempty intersection. If it was empty, then U could be separated and hence not connected.

Now there is a path between any two pair of balls, which can be shown to imply that there is a path between any two points in the union of open balls.

It is very informal but do you guys think it could work? If not, any tips on how I find the set of all points in U connected to x0 by a path in U as the hint suggests?

Last edited: Jul 29, 2013
2. Jul 29, 2013

### WannabeNewton

"And U is connected, which implies that the intersection of any two balls, can not be empty"; this is not true I'm afraid. Certainly two open balls in the union can be disjoint even if the set is connected; the set being connected just means that it cannot be written as the disjoint union of specifically two open sets.

Go with the given hint. Let $S_{0}$ be the set defined in the hint and let $x\in S_0$. $U$ is open so there exists an open ball $B_{\epsilon}(x)$ contained entirely in $U$. Use the fact that $B_{\epsilon}(x)$ is convex to show $S_0$ must be open. Then all you have to do is show that $S_0$ is closed as well (use a similar argument).

3. Jul 29, 2013

### pasmith

First show that an open ball in $\mathbb{R}^2$ is path connected.

Let $x_0 \in U$, and let $A \subset U$ be the set of points in U which can be joined to $x_0$ by a path in U.

Given that U is open, and that open balls are path connected, why does it follow that:
(1) A is not empty.
(2) A is open.
(3) A is closed, ie. the complement of A in U is open.

4. Jul 29, 2013

Thanks for the replies!

Before I post my reply, how do I write math? To the right it says "Click the sigma symbol in the toolbar for complex equations" but I can't find the sigma-symbol.

5. Jul 29, 2013

### micromass

6. Jul 29, 2013

Thank you micromass.

WannabeNewton: I found my error and changed it from "And U is connected, which implies that the intersection of any two balls, can not be empty" to "And U is connected, which implies that given any ball, there is another ball with nonempty intersection"

But I just realized why this is wrong too. Just to illustrate the error, one can have three open balls in a "chain" (the audi logo) and three other open balls in a "chain" (another audi logo), with the chains disjoint, but yet every ball intersects some other ball.

pasmith:

Open balls in $\mathbb{R}^2$ are path connected since they are convex.

1) $A$ is nonempty, since $x_0 \in A$, and also some open ball containing $x_0$.

2) It is open since it is a union of open sets (all the balls)?

3) Now the closedness of A is where I get stuck. I already know that A's complement is the empty set, but I still need to show it is open. Where do I start? By assuming the complement is the set of all numbers with no path between them? By the way, if I did not know that this in fact was the empty set, could this complement ever contain anything else than a finite number of points? Is this what I need to use? It would at least be a closed set.

Thanks again for the replies!

7. Jul 29, 2013

### WannabeNewton

Your argument for $(2)$ doesn't really prove anything. Why should it be a union of open balls? This is what you have to actually show. See what I said above in post #2. $(3)$ will become clear if you can show $(2)$.

8. Jul 30, 2013

Thank you again.

I believe I have got it right now.

Let $S_0$ be the set of points which can be joined to $x_0 \in U$ by a path in $U$. Now for any $y \in S_0$ there is some open ball $B_{\epsilon}(y)$ fully contained in $S_0$. This is true because by our definition of $S_0$, there is path between $x_0$ and $y$, and we can find a path between $x_0$ and any point in the ball too, since the ball is convex (putting the path between $x_0$ and $y$ together with the straight line segment between $y$ and any point in the ball). So for every point $y \in S_0$, there is an open set (the ball) fully contained in $S_0$, hence it is open.

We use a similir argument to show that the complement is open. Take some $y \in U\setminus S_0$. Now there is some open ball $B_{\epsilon}(y)$ fully contained in $U \setminus S_0$. By the definition of the complement, there is no path between $x_0$ and $y$, and therefore no path between the points in the ball (if there were, we could find a path between $x_0$ and $y$ too). So the complement is open too!

Thanks for all the replies.

I believe my next problem is a generalization of this problem, instead of $\mathbb{R}^2$ the space is just locally path connected.

9. Jul 30, 2013

### micromass

Seems ok!

10. Jul 30, 2013

### pasmith

On reflection, there is another proof which doesn't require showing expressly that $U \setminus S_0$ is open:

"Can be joined by a path lying wholly in U to" is an equivalence relation on U (the proof of this is straightforward). To show that U is path connected, one must show that there is exactly one equivalence class.

Since U is open, by definition for all $x \in U$ there exists $r > 0$ such that $B_r(x) \subset U$, and because $B_r(x)$ is path connected we have $B_r(x) \subset [x]$. Thus every equivalence class is open.

Since U is open, if there are at least two equivalence classes then U is the union of at least two disjoint non-empty open sets, and so by definition U is disconnected. But by assumption U is connected. Thus there is exactly one equivalence class, as required.