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Math Help - Real analysis proof

  1. #1
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    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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by quiney View Post
    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.
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  3. #3
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    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?
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