1. ## Real analysis proof

Define the relation ⊑ on converging sequences such that

α ⊑ β iff for all rationals p > 0, there is a natural number n such that for every k ≥ n, β(k) - α(k) < p. Prove that ⊑ is a total order.

I tried proving it this way:

Assume that ¬(α ⊑ β) and ¬(β ⊑ α) Thus, there is a p1 > 0, for all n ∈ Ν, there is a j ≥ n such that β(j) – α(j) ≥ p1, and there is a p2 > 0 such that for all n ∈ Ν, there is a j ≥ n such that α(j) – β(j) ≥ p2. Add these two inequalities to get 0 ≥ p1 + p2. However, since p1 and p2 are both positive, this is impossible.

I'm sure this is wrong since there is no guarantee that it's the same j for both inequalities. How might I fix this?

2. Originally Posted by quiney
Define the relation ⊑ on converging sequences such that

α ⊑ β iff for all rationals p > 0, there is a natural number n such that for every k ≥ n, β(k) - α(k) < p. Prove that ⊑ is a total order.

I tried proving it this way:

Assume that ¬(α ⊑ β) and ¬(β ⊑ α) Thus, there is a p1 > 0, for all n ∈ Ν, there is a j ≥ n such that β(j) – α(j) ≥ p1, and there is a p2 > 0 such that for all n ∈ Ν, there is a j ≥ n such that α(j) – β(j) ≥ p2. Add these two inequalities to get 0 ≥ p1 + p2. However, since p1 and p2 are both positive, this is impossible.

I'm sure this is wrong since there is no guarantee that it's the same j for both inequalities. How might I fix this?
Are you sure this right? For example, it's clearly true that if $\alpha,\beta\to q\in\mathbb{Q}$ then by your definition $\alpha\sqsubseteq \beta$ and $\beta\sqsubseteq \alpha$ yet it's evidently true that (in general) $\alpha\ne\beta$.

3. Thanks! Would it work if I switched the inequality, so that

α ⊑ β iff for all rationals p > 0, there is a natural number n such that for every k ≥ n, β(k) - α(k) > p.

But I still can't figure out how to prove the totality of ⊑. Any thoughts?