Behaviour of ($\overline{a}_n)_{n=1}^{\infty}$ and ($\underline{a}_n)_{n=1}^{\infty}$

Intuitively, you can think of $\limsup$ as doing surgery on a sequence $(a_{n})$ in this way: we start by taking $\sup$ of $(a_{n})_{n \geq 1}$; then delete $a_{1}$ and take $\sup$ of the remaining sequence, i.e. take $\sup$ of $(a_{n})_{n \geq 2}$; delete $a_{2}$ and take $\sup$ of the remaining sequence, i.e. take $\sup$ of $(a_{n})_{n \geq 3}$; repeat this procedure infinitely many times.

From the above we see that, for every $N \geq 1$, the $N$th operation on $(a_{n})$ gives exactly one number $\sup_{n \geq N}a_{n}$. Clearly, if $N_{1} < N_{2}$, then $(a_{n})_{n \geq N_{1}}$ has more components than $(a_{n})_{n \geq N_{2}}$; hence we have $\sup_{n \geq N_{1}}a_{n} \geq \sup_{n \geq N_{2}}a_{n}$.

Can you figure out how $\liminf$ works now? It works in the same fashion as how $\limsup$ works.

If $A\subset B$ then $\sup A\leq \sup B$ and $\inf A\geq \inf B$. For example if $A=\{2,1,1/2,1/4,...\}$ and $B=\{4,2,1,1/2,1/4,...\}$ then $\sup A=2<4=\sup B$.......If $A\subset B$ then any upper bound for $B$ is also an upper bound for $A$.So the least of the upper bounds for $B$ is one of the upper bounds for $A$ but maybe not the least of them.Similarly for lower bounds.

Also check out this image from Wikipedia's Limit Superior and Limit Inferior page. I think it may impart some intuition. Note how the top red line (sup) decreases while the bottom (inf) increases.