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Math Help - collection of pairwise disjoint interval question

  1. #1
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    collection of pairwise disjoint interval question

    Show that any collection of pairwise disjoint, nonempty open interval in R is at most countable. [Hint: Each one contains a rational]
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    Re: collection of pairwise disjoint interval question

    Quote Originally Posted by wopashui View Post
    Show that any collection of pairwise disjoint, nonempty open interval in R is at most countable. [Hint: Each one contains a rational]
    The hint basically gives it away. Let \left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} denote the set of your open intervals. For each I_\alpha since \mathbb{Q} is dense, we know there exists some rational number q_\alpha\in I_\alpha. So, define f:\left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} \to\mathbb{Q}:I_\alpha \mapsto q_\alpha, since the I_\alpha's are disjoint we know that f is an injection...so.
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    Re: collection of pairwise disjoint interval question

    Quote Originally Posted by Drexel28 View Post
    The hint basically gives it away. Let \left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} denote the set of your open intervals. For each I_\alpha since \mathbb{Q} is dense, we know there exists some rational number q_\alpha\in I_\alpha. So, define f:\left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} \to\mathbb{Q}:I_\alpha \mapsto q_\alpha, since the I_\alpha's are disjoint we know that f is an injection...so.
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    hi, Drexel28, what is dense means, and I did not understand the term at most countable, what is the approach here?
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    Re: collection of pairwise disjoint interval question

    dense means that between any two real numbers, we can find a rational number.

    "at most countable" means that there is an injection from our collection, to a subset of the natural numbers (this subset could be finite, or infinite).
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    Re: collection of pairwise disjoint interval question

    sorry, what is A in your approach, you have not defined A.
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    Re: collection of pairwise disjoint interval question

    Quote Originally Posted by Drexel28 View Post
    The hint basically gives it away. Let \left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} denote the set of your open intervals. For each I_\alpha since \mathbb{Q} is dense, we know there exists some rational number q_\alpha\in I_\alpha. So, define f:\left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} \to\mathbb{Q}:I_\alpha \mapsto q_\alpha, since the I_\alpha's are disjoint we know that f is an injection...so.
    _
    f:\left\{I_\alpha\right\}_{\alpha\in\mathcal{A}} \to\mathbb{Q}:I_\alpha \mapsto q_\alpha is this saying that I_a is a bijection of Q, what is the notation of I_a --> q_a mean?, I'm kindof confused here.
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    Re: collection of pairwise disjoint interval question

    what drexel28 is saying, is that A is an "indexing set" for our collection of intervals. this is mostly a notational convenience, so that we can distinguish different intervals from each other while still using a notation that tells us they all belong "to the same collection".

    since our intervals are pair-wise disjoint, one possible choice for A is the set of all real numbers {a: a = inf(I), for some interval I in our collection} (so if one of our intervals was (2,4), we would label that interval I_2).

    by your hint, each interval I_\alpha, contains a rational number, which we can call q_\alpha . since the intervals are pair-wise disjoint, no other interval contains this rational number.

    but the rationals can be put into a bijection with the natural numbers, let's call the bijection h.

    thus if C is our collection of intervals, we have an injection k from C into the natural numbers given by:

    k(I_\alpha) = h(q_\alpha)
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    Re: collection of pairwise disjoint interval question

    thx, so why do we need C here, is h(q_a) stand for nautral number?
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    Re: collection of pairwise disjoint interval question

    C is just a letter i picked to stand for \{I_\alpha\}_{\alpha \in \mathcal{A}}
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