So here's one question which I came across some time ago.
Given any sequence $a_n$ in $\mathbb{R}$, I need to show that $\lim\sup a_n\leq\sup a_n$.
My approach: As the sequence $a_n\in\mathbb{R}$, thus to talk about the $\lim\sup a_n$, I assumed that $a_n$ is bounded. I defined
$$b_n:=\sup\{a_k:k\geq n\}$$ i.e.
$$\begin{align}b_1&=\sup\{a_1,a_2,…\}\\b_2&=\sup\{a_2,a_3,…\}\\b_3&=\sup\{a_3,a_4,…\}\\.&\\.\end{align}$$
Now, I can say that $<b_n>$ is a decreasing sequence. Thus, we have, $b_1\geq b_2\geq b_3\geq …$ and so on.
Now, like we had defined $b_n$, we get that, $\sup a_n=b_1$.
I then did this:
$$\sup b_1=\sup\{b_1,b_2,…\}\geq\inf\{b_1,b_2,…\}\implies\sup a_n\geq\lim\sup a_n$$
The reason why I did this is because $<b_n>$ is a decreasing sequence. But I am a bit doubtful if I can write $\sup b_1=\sup\{b_1,b_2,…\}$. Can anyone please verify this?
Also, is there any alternate way of proving this inequality? Any help would be very much appreciable.
Best Answer
For an easier proof : For all $n\in\mathbb N$,$$\sup_{k\geq n}a_k\leq \sup_{m\in\mathbb N}a_m.$$ taking $n\to \infty $ yields the wished result.