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

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 $$$$ is a decreasing sequence. Thus, we have, $$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 $$$$ is a decreasing sequence. But I am a bit doubtful if I can write $$\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.

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.

• You're using the fact that $a_k:k\geq n$ is a subset of $a_m:m\in\mathbb{N}$? Nov 16 '20 at 11:12
• yes @DebarthaPaul
– Surb
Nov 16 '20 at 11:22
• nice one this is... Nov 16 '20 at 13:16

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 $$\lim \sup a_{m} < \sup a_{m}$$.