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The corollary 5.4.10 says, let $(a_n)_{i=0}^\infty$ and $(b_n)_{i=0}^\infty$ be Cauchy sequences of rationals such that $a_n$ >= $b_n$ for all n >= 1. Them LIM $a_n$ >= LIM $b_n$. The remark then says the corollary is not true if we replace the sign >= by >, and provides an example about two Cauchy sequences with $a_n$ := 1 + $\frac{1}{n}$ and $b_n$ := 1 - $\frac{1}{n}$, i.e. the difference between two corresponding elements is always positive, but the two corresponding LIM are equal.

Now, given the two Cauchy sequences in the remark, LIM($a_n$ - $b_n$) = LIM($a_n$) - LIM($b_n$) by the definition of subtraction of reals of the book. The real is positive since the corresponding sequence is positively bounded away from zero by definition of sequence that is positively bounded away from zero, which implies the corollary 5.4.10 applies even if we replace the sign >= with >.

I assume my reasoning has a flaw, but cannot point out where it is. Could someone explain why I am wrong?

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    $\begingroup$ Your sentence "The real is positive since the corresponding sequence is positively bounded away from zero by definition of sequence that is positively bounded away from zero" is poorly constructed. Do you mean that since $a_n-b_n$ is bounded away from zero, its limit must be positive? But $a_n-b_n=2/n$ which is not bounded away from zero in any reasonable sense of the phrase. This sequence converges to zero, which is exactly Tao's point. $\endgroup$ – Ehsaan Apr 12 at 20:37
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No, there is no contradiction here - the issue seems to be a confusion around the term "bounded away from zero."

Just because all terms of a sequence are positive does not mean that that sequence is bounded away from zero. A sequence $(x_i)_{i\in\mathbb{N}}$ is bounded away from zero iff there is some $\epsilon>0$ such that $$\vert x_i\vert>\epsilon$$ for every $i$ - that is, iff the terms of the sequence don't get "arbitrarily close" to zero.

And it's easy to check that the sequence in your question is not bounded away form zero, since for all $\epsilon>0$ there is some $n\in\mathbb{N}$ such that ${2\over n}<\epsilon$.

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