Hey guys,
I'm having a little trouble with these two questions. It would be great if you could lend me a hand
p1.
By the archimedian property, there is some n N such that,
Therefore,
(1,5)
I don't understand why
Also why does (1,5) ?
Hey guys,
I'm having a little trouble with these two questions. It would be great if you could lend me a hand
p1.
By the archimedian property, there is some n N such that,
Therefore,
(1,5)
I don't understand why
Also why does (1,5) ?
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.
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].
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].