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

  1. Jul 29, 2013 #1
    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?

    Thanks in advance!
     
    Last edited: Jul 29, 2013
  2. jcsd
  3. Jul 29, 2013 #2

    WannabeNewton

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    "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).
     
  4. Jul 29, 2013 #3

    pasmith

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    First show that an open ball in [itex]\mathbb{R}^2[/itex] is path connected.

    Let [itex]x_0 \in U[/itex], and let [itex]A \subset U[/itex] be the set of points in U which can be joined to [itex]x_0[/itex] 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.
     
  5. Jul 29, 2013 #4
    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.
     
  6. Jul 29, 2013 #5

    micromass

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  7. Jul 29, 2013 #6
    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 [itex] \mathbb{R}^2 [/itex] are path connected since they are convex.

    1) [itex] A [/itex] is nonempty, since [itex] x_0 \in A [/itex], and also some open ball containing [itex] x_0 [/itex].

    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!
     
  8. Jul 29, 2013 #7

    WannabeNewton

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    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)##.
     
  9. Jul 30, 2013 #8
    Thank you again.

    I believe I have got it right now.

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

    We use a similir argument to show that the complement is open. Take some [itex]y \in U\setminus S_0 [/itex]. Now there is some open ball [itex] B_{\epsilon}(y) [/itex] fully contained in [itex] U \setminus S_0 [/itex]. By the definition of the complement, there is no path between [itex]x_0 [/itex] and [itex]y [/itex], and therefore no path between the points in the ball (if there were, we could find a path between [itex]x_0 [/itex] and [itex] y[/itex] 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 [itex]\mathbb{R}^2 [/itex] the space is just locally path connected.
     
  10. Jul 30, 2013 #9

    micromass

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    Seems ok!
     
  11. Jul 30, 2013 #10

    pasmith

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    On reflection, there is another proof which doesn't require showing expressly that [itex]U \setminus S_0[/itex] 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 [itex]x \in U[/itex] there exists [itex]r > 0[/itex] such that [itex]B_r(x) \subset U[/itex], and because [itex]B_r(x)[/itex] is path connected we have [itex]B_r(x) \subset [x][/itex]. 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.
     
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