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Math Help - Archimedian Property Question

  1. #1
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    Archimedian Property Question

    Hey guys,

    I'm having a little trouble with these two questions. It would be great if you could lend me a hand

    p1.

     \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] = (1 , 5)

    By the archimedian property, there is some n  \in N such that,  \frac{1}{n} < \epsilon

    Therefore,  \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] \geq \cup_{k=1}^{\infty} [1 +\epsilon, 5 - \epsilon] \geq (1,5)

    (1,5)  \geq \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] = (1 , 5)

    I don't understand why  \cup_{k=1}^{\infty} [1 +\epsilon, 5 - \epsilon] \geq (1,5)

    Also why does (1,5)  \geq \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] = (1 , 5) ?
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  2. #2
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    Re: Archimedian Property Question

    I would do this problem as follows. Let x ∈ (1, 5) and let e = min(x - 1, 5 - x) > 0. Then there exists an n such that 1/n < e. Therefore, x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪k=1[1 + 1/k, 5 - 1/k]. The converse inclusion is obvious because [1 + 1/k, 5 - 1/k] ⊆ (1, 5) for every k.
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  3. #3
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    Re: Archimedian Property Question

    Why did you set e as the minimum of (x-1 , 5-x)?

    Also can you elaborate on how x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k] shows that (1,5) ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k]?
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  4. #4
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    Re: Archimedian Property Question

    Quote Originally Posted by rokman54 View Post
    p1.
     \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] = (1 , 5)
    By the archimedian property, there is some n  \in N such that,  \frac{1}{n} < \epsilon
    Is it clear to you that (\forall k)~[1 +\frac{1}{k}, 5 - \frac{1}{k}]\subset (1,5)~?

    Suppose that t\in (1,5). Now use the Archimedian property.
    \exists K such that t\in [1 +\frac{1}{K}, 5 - \frac{1}{K}].

    That means (1,5)\subseteq \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}]
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  5. #5
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    Re: Archimedian Property Question

    Quote Originally Posted by rokman54 View Post
    Why did you set e as the minimum of (x-1 , 5-x)?
    Because then e ≤ x - 1 and e ≤ 5 - x. The first inequality implies that 1 + e ≤ x, and the second one implies that x ≤ 5 - e. Therefore, x ∈ [1 + e, 5 - e]. But since 1/n ≤ e, it follows that 1 + 1/n ≤ 1 + e and 5 - e ≤ 5 - 1/n. Thus, [1 + e, 5 - e] ⊆[1 + 1/n, 5 - 1/n].

    Quote Originally Posted by rokman54 View Post
    Also can you elaborate on how x ∈ [1 + e, 5 - e] ⊆ [1 + 1/n, 5 - 1/n] ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k] shows that (1,5) ⊆ ∪k=1∞[1 + 1/k, 5 - 1/k]?
    I started by assuming x ∈ (1, 5) and concluded that x ∈ ∪k=1[1 + 1/k, 5 - 1/k]. This shows that (1, 5) ⊆ ∪k=1[1 + 1/k, 5 - 1/k].
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