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Math Help - Connectedness

  1. #1
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    Connectedness

    Consider the two tangent open discs \{x^2+y^2<1\} and \{(x-2)^2+y^2<1\} in \mathbb{R}^2.
    How can show that the union of one disc and the closure of the other is a connected subset of \mathbb{R}^2?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Use the following proposition:

    If A and B are connected sets which are not separated, then A\cup B is connected.

    Choose A an open disc and B the closure of the other one.

    Fernando Revilla
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Use the following proposition:

    If A and B are connected sets which are not separated, then A\cup B is connected.

    Choose A an open disc and B the closure of the other one.

    Fernando Revilla
    Don't forget to mention that the closure of connected subspaces is connected.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Don't forget to mention that the closure of connected subspaces is connected.
    Don't forget that a hint is not a complete solution of a problem.

    Fernando Revilla
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  5. #5
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    Thanks all. I have another solution using this theorem: If E is a connected subset of X and E\subset A\subset\overline{E}, then A is connected.

    We have X=E\cup\overline{F}. Since E is a connected subset of X and E\subset A=(\{x^2+y^2<1\}\cup(0,-1))\subset\overline{E} it follows that A is also connected. Thus A and \overline{F} are connected subsets of X such that A\cap\overline{F}=(0,-1)\neq\emptyset. Then it follows that A\cup\overline{F}$=$E\cup\overline{F}=X is connected.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    That is right, you have used:

    Proposition

    Let \mathcal{A}=\{A_i\} be a class of connected subsets of X (toplogical space) with non-empty intersection. Then \big\cup_i A_i is connected.

    And this proposition is as a consequence that any two members of \mathcal{A} are not separated. All depends on the starting point.

    Fernando Revilla
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