# Tao Lemma 6.4.13.

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.

• 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}". – xbh Jun 16 '19 at 3:07
• Thanks, I appreciate it. I'll have to fix that. – user465188 Jun 16 '19 at 3:17
• How do you get that $b_n$ is an upper bound for $(a_n)_{n=m}^{\infty}.$? – WhoKnowsWho Jun 16 '19 at 3:27
• I thought that this followed from the assumption that $a_n \leq b_n$ for all $n \geq m$. – user465188 Jun 16 '19 at 3:28
• 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. – DanielWainfleet Jun 16 '19 at 3:30

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.