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Thread: Proof that union of two connected non disjoint sets is connected

  1. #1
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    Proof that union of two connected non disjoint sets is connected

    The following is an attempt at a proof which I wrote up for a homework problem for Advanced Calc. I got some help from my professor briefly the morning it was due, but I'm not quite sure if I grasped exactly what he was saying. I think he said that somehow A_(1) instersects both S_(1) and A_(2) intersects S_(2)... Sorry about the formatting. ( I have written papers in LaTeX but I dont know exactly how to format properly with this GUI editor.) Please show me what I did wrong, because I know I was definitely on the right track with this proof.


    Suppose S_(1) and S_(2) are connected non disjoint sets.
    Show that S_(1) U S_(2) is connected.

    Proof by contradiction:
    Assume S_(1) U S_(2) is disconnected.

    Then there exists A_(1) and A_(2) such that

    1. A_(1) U A_(2) = S_(1) U S_(2)
    2. closure {A_(1)} \int A_(2) = closure {A_(2)} \int A_(1)

    Case 1: A_(1) and A(2) cannot be S_(1) and S_(2) or S_(2) and S_(1)
    respectively.

    If either of these cases is true then 2. is violated.

    We can then construct S_(1) or S_(2) from a piece of A_(1) and A_(2).
    These two pieces then do not intersect by 1.
    Then S_(1) or S_(2) is the union of non empty disjoint sets and hence it is disconnected.

    ----><----- contradiction.

    So S_(1) U S_(2) is connected.
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  2. #2
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    Quote Originally Posted by csuMath&Compsci View Post
    The following is an attempt at a proof which I wrote up for a homework problem for Advanced Calc. I got some help from my professor briefly the morning it was due, but I'm not quite sure if I grasped exactly what he was saying. I think he said that somehow A_(1) instersects both S_(1) and A_(2) intersects S_(2)... Sorry about the formatting. ( I have written papers in LaTeX but I dont know exactly how to format properly with this GUI editor.) Please show me what I did wrong, because I know I was definitely on the right track with this proof.


    Suppose S_(1) and S_(2) are connected non disjoint sets.
    Show that S_(1) U S_(2) is connected.

    Proof by contradiction:
    Assume S_(1) U S_(2) is disconnected.

    Then there exists A_(1) and A_(2) such that

    1. A_(1) U A_(2) = S_(1) U S_(2)
    2. closure {A_(1)} \int A_(2) = closure {A_(2)} \int A_(1) I think you mean $\displaystyle \color{red}\overline{A_1}\cap A_2 = \overline{A_2}\cap A_1 = \emptyset.$

    Case 1: A_(1) and A(2) cannot be S_(1) and S_(2) or S_(2) and S_(1)
    respectively.

    If either of these cases is true then 2. is violated.

    We can then construct S_(1) or S_(2) from a piece of A_(1) and A_(2).
    These two pieces then do not intersect by 1.
    Then S_(1) or S_(2) is the union of non empty disjoint sets and hence it is disconnected.

    ----><----- contradiction.

    So S_(1) U S_(2) is connected.
    Your proof is along the right lines, but you haven't quite tied it together. You're assuming that $\displaystyle S_1\cup S_2$ is disconnected. That means that the sets $\displaystyle A_1$ and $\displaystyle A_2$ satisfy not only 1. and 2., but also

    $\displaystyle 3.\quad (S_1\cup S_2)\cap A_1\ne\emptyset \text{ and } (S_1\cup S_2)\cap A_2\ne\emptyset.$

    Now you can argue as follows. It's not possible for $\displaystyle S_1\cap A_1$ and $\displaystyle S_1\cap A_2$ both to be nonempty, because $\displaystyle S_1$ is connected. So the whole of $\displaystyle S_1$ must lie in one or other of $\displaystyle A_1$ and $\displaystyle A_2$, say $\displaystyle S_1\subseteq A_1$.

    Similarly, we must have $\displaystyle S_2\subseteq A_1$ or $\displaystyle S_2\subseteq A_2$. But if $\displaystyle S_2\subseteq A_1$ then $\displaystyle (S_1\cup S_2)\subseteq A_1$, which means that $\displaystyle (S_1\cup S_2)\cap A_2$ is empty, contradicting 3.

    Therefore $\displaystyle S_2\subseteq A_2$. But we are told that $\displaystyle S_1\cap S_2\ne\emptyset$. So choose $\displaystyle x\in S_1\cap S_2$. Then $\displaystyle x\in S_1\subseteq A_1$, and $\displaystyle x\in S_2\subseteq A_2$. Therefore $\displaystyle A_1\cap A_2\ne\emptyset$. But that contradicts 2. So either way we have a contradiction.
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