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Thread: Connected sets

  1. #1
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    Connected sets

    Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.

    Hint: (Use) Let G∪H be a disconnection of A and let B be a connected subset of A then we see that either BH= or B∩G=∅, and so either BG or B⊂H.
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  2. #2
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    Quote Originally Posted by AcCeylan View Post
    Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.
    Suppose that $\displaystyle G\cup H=B$ is a disconnection of $\displaystyle B$.

    That means that $\displaystyle G\cap B\ne\emptyset~\&~H\cap B\ne\emptyset$.

    Also $\displaystyle G\cap\overline{H}=\emptyset~\&~H\cap\overline{G}=\ emptyset$

    Using the fact that $\displaystyle B\subseteq\overline{A}$ prove that contradicts the given that $\displaystyle A$ is connected.
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    Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
    if not how can I do this , can you explain, please thank you
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  4. #4
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    Quote Originally Posted by AcCeylan View Post
    Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
    THEOREM The closure of a connected set is a connect set.
    That is what you are really asked to prove.

    As a lemma (previous theorem) you should have proved that if $\displaystyle A$ is a connected set and $\displaystyle A\subset G\cup H$ where $\displaystyle G~\&~H$ are separated sets then $\displaystyle A\subset G$ or $\displaystyle A\subset H $.
    Two set are said to be separated if neither is empty and neither contains a point nor a limit of the other.

    Now it is quite easy to prove the lemma. If the conclusion is false then $\displaystyle (A\cap G)\cup(A\cap H)=A$ is a separation of $\displaystyle A$.
    But $\displaystyle A$ is connected. So that is a contradiction.
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  5. #5
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    Thanks for everything but I have one more question..you said that
    "Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
    How can I show this?
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  6. #6
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    Quote Originally Posted by AcCeylan View Post
    Thanks for everything but I have one more question..you said that
    "Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
    How can I show this?
    Suppose that $\displaystyle G\cup H=B$ is a separation of $\displaystyle B$.
    Because $\displaystyle A\subseteq B$ and $\displaystyle A$ is connected we have $\displaystyle A\subseteq G$ or $\displaystyle A\subseteq B$.
    Say $\displaystyle A\subseteq G$. Now is it possible for $\displaystyle H$ to be nonempty?
    If the answer is no, then you have a contradiction. And that means that $\displaystyle B$ is connected.
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