text: # Inequalities with Cauchy sequences.
  
  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 $\forall n\geq 1,a_n\leq x<q$, construct a sequence $(q−an)$, it's trivial to prove the sequence is Cauchy and it's non-negative...
  
  I don't understand why is $(q−a_n)$ non-negative, because, $$a_n\leq x<q, \qquad n\geq 1,$$ $$\Rightarrow a_n<q ,$$ $$\Rightarrow 0<(q-a_n).$$
  
  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,
  
  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 "$>$" ?
  
  • 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? – Juniven Jan 20 '17 at 18:05
  • Just like $3>2$ and so there is no problem by writing $3\geq 2$. – Juniven Jan 20 '17 at 18:07
  • 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. – Keith Jan 20 '17 at 18:26
explanation: The text extract is a discussion about a proof involving Cauchy sequences and inequalities. The user is asking for clarification on two specific points in the proof:
  
  1. Why the sequence \((q - a_n)\) is non-negative.
  2. The use of inequality signs (\(\geq\) and \(>\)) in a proof from Terence Tao's "Analysis I" regarding the existence of the least upper bound.
  
  The user provides their own reasoning for the first point and seeks confirmation. For the second point, the user is confused about the transition from a strict inequality to a non-strict inequality.
  
  The text contains a question and an answer to the first point, but the second point is left as a question without an answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Inequalities with Cauchy sequences.
  
  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 $\forall n\geq 1,a_n\leq x<q$, construct a sequence $(q−an)$, it's trivial to prove the sequence is Cauchy and it's non-negative...
  
  I don't understand why is $(q−a_n)$ non-negative, because, $$a_n\leq x<q, \qquad n\geq 1,$$ $$\Rightarrow a_n<q ,$$ $$\Rightarrow 0<(q-a_n).$$
  
  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,
  
  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 "$>$" ?
  
  • 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? – Juniven Jan 20 '17 at 18:05
  • Just like $3>2$ and so there is no problem by writing $3\geq 2$. – Juniven Jan 20 '17 at 18:07
  • 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. – Keith Jan 20 '17 at 18:26
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion about a proof involving Cauchy sequences and inequalities. The user is asking for clarification on two specific points in the proof:
  
  1. Why the sequence \((q - a_n)\) is non-negative.
  2. The use of inequality signs (\(\geq\) and \(>\)) in a proof from Terence Tao's "Analysis I" regarding the existence of the least upper bound.
  
  The user provides their own reasoning for the first point and seeks confirmation. For the second point, the user is confused about the transition from a strict inequality to a non-strict inequality.
  
  The text contains a question and an answer to the first point, but the second point is left as a question without an answer.
  
  Label: question-and-answer, question