collection of pairwise disjoint interval question

**wopashui** · October 10th 2011, 06:16 PM

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

**Drexel28** · October 10th 2011, 07:49 PM

Originally Posted by wopashui

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.
_

**wopashui** · October 11th 2011, 05:38 PM

Originally Posted by Drexel28

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.
_

hi, Drexel28, what is dense means, and I did not understand the term at most countable, what is the approach here?

**Deveno** · October 11th 2011, 05:50 PM

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).

**wopashui** · October 11th 2011, 05:53 PM

sorry, what is A in your approach, you have not defined A.

**wopashui** · October 11th 2011, 06:07 PM

Originally Posted by Drexel28

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.

**Deveno** · October 11th 2011, 07:02 PM

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:

$k(I_\alpha) = h(q_\alpha)$

**wopashui** · October 11th 2011, 07:27 PM

thx, so why do we need C here, is $h(q_a)$ stand for nautral number?

**Deveno** · October 11th 2011, 07:35 PM

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

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