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I am trying to prove this following lemma in Tao's analysis text.

Suppose that $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are two sequences of real numbers such that $a_n \leq b_n$ for all $n \geq m$. Then we have the inequalities: \begin{align*} & (a) \sup(a_n)_{n=m}^{\infty} \leq \sup (b_n)_{n=m}^{\infty} \\ & (b) \inf(a_n)_{n=m}^{\infty} \leq \inf(b_n)_{n=m}^{\infty} \\ & (c) \lim \sup_{n \to \infty} a_n \leq \lim \sup_{n \to \infty} b_n \\ & (d) \lim \inf_{n \to \infty} a_n \leq \lim \inf_{n \to \infty} b_n. \end{align*}

Here is my attempt.

(a) We have, by the definition of the supremum, that for any $n \geq m$, \begin{align*} & a_n \leq \sup(a_n)_{n=m}^{\infty} \\ & b_n \leq \sup(b_n)_{n=m}^{\infty}. \end{align*} Since $b_n$ is an upper bound of $(a_n)_{n=m}^{\infty}$, it must be no greater than $\sup (a_n)_{n=m}^{\infty}$. Thus, \begin{align*} a_n \leq \sup(a_n)_{n=m}^{\infty} \leq b_n \leq \sup(b_n)_{n=m}^{\infty}, \end{align*} and this means that $\sup(a_n)_{n=m}^{\infty} \leq \sup(a_n)_{n=m}^{\infty}$.

(b) By the definition of infimum, for any $n \geq m$, we have \begin{align*} & a_n \geq \inf (a_n)_{n=m}^{\infty} \\ & b_n \geq \inf (b_n)_{n=m}^{\infty} \end{align*} Since $a_n$ is a lower bound for $(b_n)_{n=m}^{\infty}$, it must be no larger than the infimum of $(b_n)_{n=m}^{\infty}$, so \begin{align*} b_n \geq \inf (b_n)_{n=m}^{\infty} \leq a_n \geq \inf (a_n)_{n=m}^{\infty}, \end{align*} and thus $\inf (a_n)_{n=m}^{\infty} \leq \inf (b_n)_{n=m}^{\infty}$.

I a bit stumped on (c) and (d), and am unsure on whether I've made progress. For (c), I so far have used Proposition 6.4.12, part (c), in Tao's text to say that \begin{align*} \inf (a_n)_{n=m}^{\infty} \leq \lim \inf a_n \leq \lim \sup a_n \leq \sup (a_n)_{n=m}^{\infty} \end{align*} and, similarly, that \begin{align*} \inf (b_n)_{n=m}^{\infty} \leq \lim \inf b_n \leq \lim \sup b_n \leq \sup (b_n)_{n=m}^{\infty}. \end{align*} By part (a), we have $\inf (a_n)_{n=m}^{\infty} \leq \inf (b_n)_{n=m}^{\infty}$, so \begin{align*} \inf (a_n)_{n=m}^{\infty} \leq \inf (b_n)_{n=m}^{\infty} \leq \lim \inf b_n \leq \lim \sup b_n \leq \sup (b_n)_{n=m}^{\infty}, \end{align*} hence, \begin{align*} \inf (a_n)_{n=m}^{\infty} \leq \lim \sup (b_n)_{n=m}^{\infty}. \end{align*} It feels that I've made progress because I have been able to derive a statement that includes both sequences. But, I am still not quite how to introduce the $\lim \inf$ into this expression.

Any help would be greatly appreciated.

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    $\begingroup$ Typesetting tip: "\limsup_{n\to \infty}" generates $$ \limsup_{n\to \infty},$$ which is better [as to me] than $$ \lim\sup_{n\to \infty}$$ that is generated by "\lim\sup _{n \to\infty}". $\endgroup$ – xbh Jun 16 '19 at 3:07
  • $\begingroup$ Thanks, I appreciate it. I'll have to fix that. $\endgroup$ – user465188 Jun 16 '19 at 3:17
  • $\begingroup$ How do you get that $b_n$ is an upper bound for $(a_n)_{n=m}^{\infty}.$? $\endgroup$ – WhoKnowsWho Jun 16 '19 at 3:27
  • $\begingroup$ I thought that this followed from the assumption that $a_n \leq b_n$ for all $n \geq m$. $\endgroup$ – user465188 Jun 16 '19 at 3:28
  • $\begingroup$ NO. "Since $b_n$ is an upper bound for $(a_n)_{n=m}^{\infty}$ it must be no greater than $\sup (a_n)_{n=m}^{\infty}$" is all wrong. $\endgroup$ – DanielWainfleet Jun 16 '19 at 3:30
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I do not have enough reputation to make a comment, so I will put it in the form of an answer.

In regards to the first part, you cannot say that $b_n$ is an upper bound. What you can say though is that $b_n \leq \sup(b_n)_{n=m}^{\infty}$. Thus, $a_n \leq \sup(b_n)_{n=m}^{\infty}$ for $n \geq m$ (which you have). Thus, $\sup(b_n)$ is an upper bound of $a_n$ for each $n$ and as a result, $\sup(a_n)_{n=m}^{\infty} \leq \sup(b_n)_{n=m}^{\infty}$.

In regards to your part (c), it looks like you are essentially done. While I have never read Tao's text, I would imagine that you have seen the comparison theorem. The theorem states something like if you have sequences $a_n,b_n$ such that $\lim_{n\to\infty} a_n = a$ and $\lim_{n\to\infty} b_n = b$ and $a_n \leq b_n$ for $n \geq N$, then $a \leq b$. Thus, you can take the limit as $n \rightarrow \infty$ in your last statement to get the desired result.

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