I came across the following exercise while self-studying Terrence Tao's book Analysis I:
Exercise 6.4.2 Let $(a_n)_{n=m}^\infty$ be an arbitrary sequence of real numbers and let $m'\ge m$ and $k\ge 0$ be integers. Show that $\limsup_{n\to\infty} a_n$ is equivalent to
- The limit superior of of $(a_n)_{n=m'}^\infty$.
- The limit superior of $(a_{n+k})_{n=m}^\infty$.
My Attempt: Let us first prove 1. Suppose that the limit superior of $(a_n)_{n=m}^\infty$ is equal to some real number $c\in\Bbb R$, i.e. $\inf(a_N^+)_{N=m}^\infty = c$ where $(a_N^+)_{N=m}^\infty$ denotes the sequence of all $a_N=\sup(a_n)_{n=N}^\infty$. If $M'\ge M\ge m$ then $a_{M'}^+\le a_M^+$, as if $a_M^+\le a_{M'}^+$ then $$\forall n\ge M: a_n\le a_M^+\le a_{M'}^+,$$ and $a_{M'}^+$ cannot be the lowest upper bound of $(a_n)_{n=M'}^\infty$. Thus, $(a_N^+)_{N=m}^\infty$ is decreasing. To prove that $c$ is a lower bound of $(a^+_N)_{N=m'}^\infty$, suppose that there is some $a_{n'}<c$ where $n'\le m'$. Then $$\exists n'\ge m'\forall n\ge m:a_{n'}<c\le a_n\implies \exists n'\ge m': a_{n'}<a_{n'},$$ a contradiction. Conversely, suppose that $\inf(a^+_N)_{N=m'}^\infty = c$. Then once again, $(a^+_N)_{N=m'}^\infty$ is decreasing and as such, $$\forall n'\ge m'\ge n\ge m:a_n\ge a_{n'}\ge c.$$ Hence, $\inf(a_N^+)_{N=m}^\infty = c$. For 2, note that $(a_{n+k})_{n=m}^\infty$ is precisely the same sequence as $(a_n)_{n=m+k}^\infty$, thus part 1 is applicable here since $m\le m+k$.
Is this reasoning correct, or have I done something wrong? Thanks in advance.
Edit: I realize now that it was unnecessary to prove that $(a^+_N)_{N=m}^\infty$ in the first bit of my proof of 1, even though it is applicable to the second part. Granted, one could cope with this by using a contradiction like $$\exists n'\ge m'\forall n\ge m:a_{n'}<c\le a_n\implies \exists n'\ge m': a_{n'}<a_{n'+1},$$ instead (this should imply that $(a^+_N)_{N=m}^\infty$ is not decreasing).