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.

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

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

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

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

Also why does (1,5) $\displaystyle \geq \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] = (1 , 5) $ ?

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.

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]?

Re: Archimedian Property Question

Quote:

Originally Posted by

**rokman54** p1.

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

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

Is it clear to you that $\displaystyle (\forall k)~[1 +\frac{1}{k}, 5 - \frac{1}{k}]\subset (1,5)~?$

Suppose that $\displaystyle t\in (1,5)$. Now use the Archimedian property.

$\displaystyle \exists K$ such that $\displaystyle t\in [1 +\frac{1}{K}, 5 - \frac{1}{K}]$.

That means $\displaystyle (1,5)\subseteq \cup_{k=1}^{\infty} [1 +\frac{1}{k}, 5 - \frac{1}{k}] $

Re: Archimedian Property Question

Quote:

Originally Posted by

**rokman54** 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** 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].