How to show limit of inf is less than or equal to to limit of sup

Question

I am trying to prove that the limit of a inf is less than or equal to the limit of sup for some bounded sequence $a_n$.

There are two cases, when it converges and when it doesn't. I know that when it converges the limits of sup and inf are equal which fits the theorem.

But I'm not sure what to do next. I have seen another answer on here but that defines other sequences like this: $$a_n^+=\sup\{ a_k : k>n \} $$

To me this means that we take each subsequence of $a_n$ beginning with n=1 and then n=2 etc. And then fond what the sup of that new sequence is and make that the term of our new sequence. If that is the case then that is fine but I dont understand how that can be said to be monotone on any way. As $a_n$ could be very erratic?

I cannot work through the proofs I see because I don't understand how this can be.

Answer 1

As @Dr. MV points out, the infimum of a set is a lower bound, and the supremum of a set is an upper bound, so for any set $A$ of real numbers, $\inf_{x\in A} x\le \sup_{x\in A} x$.

Define the sets $A_n=\{x_n,x_{n+1},...\}$. Then $\inf_{x\in A_n} x\le \sup_{x\in A_n} x$ for all $n$. Notice that $\{\inf A_n\}_{n\in\mathbb{N}}$ and $\{\sup A_n\}_{n\in\mathbb{N}}$ are sequences of real numbers. Since limits preserve weak inequalities, we have the result.

Answer 2

Here is another way to look at it.

Suppose $\langle a_n\rangle_{n\in\Bbb N}$ is a bounded sequence, bounded above by $M$ and below by $m,$ say. Consider the "tails" of the sequence $A_n:=\{a_k:k\ge n\}$ (as sets, rather than as subsequences). Then a few facts are readily seen:

• Each $A_n$ is infinite (so non-empty), and is bounded above by $M$ and below by $m,$ so each such $A_n$ has both a supremum and an infimum.
• For each $n\in\Bbb N,$ $A_n\supseteq A_{n+1}.$ Consequently, any upper (resp. lower) bound of $A_n$ will automatically be such a bound for $A_{n+1},$ which we can confirm by checking the definition of upper (lower) bound of $A_{n+1}.$
• In particular, $\sup A_n$ is an upper bound of $A_{n+1},$ and $\inf A_n$ is a lower bound of $A_{n+1},$ so by definition of supremum and infimum of $A_{n+1},$ we have $\sup A_n\geq\sup A_{n+1}$ and $\inf A_n\leq\inf A_{n+1}.$
• Setting $b_n:=\sup A_n$ and $c_n:=\inf A_n$ for all $n\in\Bbb N,$ we have by the above that $\langle b_n\rangle_{n\in\Bbb N}$ is a monotone nonincreasing sequence, and that $\langle c_n\rangle_{n\in\Bbb N}$ is a monotone nondecreasing sequence.
• Furthermore, noting that $m\leq a_n\leq b_n$ and $c_n\leq a_n\leq M$ for all $n\in\Bbb N,$ we have that both of these sequences are bounded, and so each has a supremum and an infimum of its own.
• Setting $b:=\inf_{n\in\Bbb N} b_n$ and $c:=\sup_{n\in\Bbb N}c_n,$ we readily have that $b=\lim_{n\to\infty}b_n$ and $c=\lim_{n\to\infty}c_n.$
• Finally, noting that $c_n\leq a_n\leq b_n$ for all $n\in\Bbb N,$ we find that $c_n\leq b_n$ for all $n.$ For "most" sequences $\langle a_n\rangle_{n\in\Bbb N}$, we can actually conclude that $c_n<b_n$ for all $n\in\Bbb N,$ and we can always draw this conclusion if $\langle a_n\rangle_{n\in\Bbb N}$ doesn't converge. (Do you see why?)
• Thus, $b_n-c_n$ is a nonnegative ("probably" positive) number for all $n\in\Bbb N,$ so since the sequences $\langle b_n\rangle_{n\in\Bbb N}$ and $\langle c_n\rangle_{n\in\Bbb N}$ both converge, then so does their difference, and $\lim_{n\in\Bbb N}(b_n-c_n)=b-c.$
• As the limit of a sequence of nonnegative numbers, we know that $b-c$ is also nonnegative (why?), so we can conclude the following (equivalent) statements: $$c\leq b,\\\sup_{n\in\Bbb N}\left(\inf A_n\right)\leq\inf_{n\in\Bbb N}\left(\sup A_n\right),\\\lim_{n\in\Bbb N}\left(\inf A_n\right)\leq\lim_{n\in\Bbb N}\left(\sup A_n\right),\\\liminf_{n\in\Bbb N}a_n\leq\limsup_{n\in\Bbb N}a_n.$$

Summed up in words, as we consider later and later "tails" of the given bounded sequence, the tails respective suprema may decrease, but will never increase; their respective infima may increase, but will never decrease. As both of these sequences are bounded (for example, by the bounds of the original sequence), then the sequences of suprema and infima of tails converge.

Since the supremum of a given "tail" cannot be less (and will "probably" be greater) than its infimum, then the limit of the sequence of suprema of the "tails" (which is the limit superior of the original sequence) cannot be less than the limit of the sequence of infima of the "tails" (which is the limit inferior of the original sequence).