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Math Help - A few proofs I need help with to get started...

  1. #1
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    Post A few proofs I need help with to get started...

    I'm really quite distraught on two particular questions... I have attempted them, but I am new at proofs and help / ideas as to how to start these two proofs would be greatly appreciated! Thank you in advance!!

    Prove that the only set in R1 which is open and closed are the empty set and R1 itself.

    Given a set S in Rn with the property that for any x that is an element of S there is an n-ball B(x) such that B(x)S is countable. Prove that S is countable.

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  2. #2
    Super Member Deadstar's Avatar
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    Here's part 1.

    This is a question from my tutorial that basically drags you through the proof. I'll post the question to let you have a go at it and if you can't answer it I'll start posting the solution! Note that the compliment of \mathbb{R} is the empty set. So if \mathbb{R} is both open and closed so is the empty set. Here's the question!


    Let A be a non-empty subset of \mathbb{R} and suppose that A is both open and closed. Fix a point a \in A and define

    B = \{ b \in \mathbb{R} : b > a \textrm{ and} [a, b) \subseteq A \}:

    Note that B is non-empty (since A is open).

    Prove that B is not bounded above. [Assume that it is and obtain contradictions both when sup B \in A, since A is open, and when sup B \notin A, since A is closed.] Deduce that [a,\infty) \subseteq A.

    By considering

    C = \{c \in \mathbb{R} : c < a \textrm{ and} (c, a] \subseteq A\}.

    prove similarly that (- \infty, a] \subseteq A (here the contradictions arise by considering inf C).
    Deduce that A = \mathbb{R}.
    We have thus shown that the only subsets of \mathbb{R} that are both open and closed are \mathbb{R} and
    the empty set.
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  3. #3
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    Thank you very much for your help! However, I'm not too sure how to prove this particular line:

    "prove similarly that (here the contradictions arise by considering inf )"

    I understand the first part though! Thank you very much!
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  4. #4
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    There is a really simple proof if you know that \mathcal{R} is connected.
    If non-empty T \varsubsetneq \mathcal{R} which is both open and closed then T^c, the complement, is non-empty and is both open and closed.

    But T \cup T^c  = \mathcal{R} which is a separation of \mathcal{R}.
    That is a contradiction.
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