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I'm reading this question (Terence Tao Exercise 5.4.8: Boundedness of Limit.) regarding the proof a property of Cauchy sequences. However, I don't understand a step in the proof,

... suppose $a = \text{LIM}_{n\rightarrow \infty}a_n >x$, there exists a rational $q$ that $a>q>x$ due to Prop 5.4.14. Since $\forall n\geq 1,a_n\leq xnon-negative...

I don't understand why is $(q−a_n)$ non-negative, because, $$a_n\leq x

Hence the sequence is strictly positive, this is critical in the next step because the following theorem is valid for non-negative sequences only.

I'm also confused with the use of the "$\geq$" and "$>$" in a few proofs of real analysis, for example in page (117,118) of Terence Tao Analysis I, when he proves the (Existence of least upper bound), he performs the following step,

Terry Tao book, (Existence of least upper bound)

How can he conclude that,

$$\Big| \frac{m_n}{n} - \frac{m_{n}}{n'} \Big|\leq \frac{1}{N} \quad \text{for all } \quad n,n' \geq N\geq 1,$$

since in a previous equation he had:

$$ \frac{m_n}{n} - \frac{m_{n}}{n'} > -\frac{1}{N} $$

Can one just put the "$\geq$" instead of the "$>$" ?

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    You already said that $$\frac{m_n}{n}-\frac{m_n}{n'}>-\frac{1}{N}.$$ So, it must be clear then that $$\frac{m_n}{n}-\frac{m_n}{n'}\geq-\frac{1}{N}$$ is a true statement. Remember the definition of the disjunction of two propositions? – 2017-01-20
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    Just like $3>2$ and so there is no problem by writing $3\geq 2$. – 2017-01-20
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    I understand, the problem is the other way around, starting from a statement where $\leq$ and claim that it is a $<$. Just as you said, the disjunction of two propositions. – 2017-01-20

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