Prove that $\lim\sup a_n\leq \sup a_n$

Question

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.

Answer 1

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.

Answer 2

The following are just some ideas, I think a similar argument works but it seems iffy/wrong:

If $\lim \sup a_{n} > \sup a_{n}$, then $\lim \sup a_{n} - \sup a_{n} > \epsilon_{1}$ for some $\epsilon_{1} > 0$.

By definition $\lim \sup a_{n} = L$ such that $|\sup a_{m} - L| < \epsilon$ for every $\epsilon > 0$ where $m \geq n$.

For this last inequality set $\epsilon := \epsilon_{1}$.

Then $|\sup a_{m} - L| < \epsilon_{1}$ and $L - \sup a_{n} > \epsilon_{1}$ so that $|\sup a_{m} - L| < L -\sup a_{m}$

By supposition we had that $L = \lim \sup a_{n} > \sup a_{n}$ so that $|\sup a_{m} - L| = L-\sup a_{m} < L -\sup a_{m}$, a contradiction.

Therefore it must be that $\lim \sup a_{m} < \sup a_{m}$.