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?

Quote:

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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?

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Are you sure this right? For example, it's clearly true that if then by your definition and yet it's evidently true that (in general) .

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?