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

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.

• The supremum is the least upper bound, while the infimum is the greatest lower bound. Hence the supremum is greater than or equal to the infimum. – Mark Viola Mar 21 '17 at 2:21
• I'm talking about the limit of the infimum and the supremum though and I need to show it rigorously – John O'Neil Mar 21 '17 at 2:31
• Yes, I understand that. So, $\limsup_n x_n=\lim_n \sup_{m\ge n}(x_m)\ge \lim_n (\inf_{m\ge n})(x_m)=\liminf_n x_n$. – Mark Viola Mar 21 '17 at 2:40
• Sorry, I misinterpreted. This topic confuses me – John O'Neil Mar 21 '17 at 3:57

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.
• Okay I believe I see what you are saying but just for clarification are those sets $\{ \sup A_n \}$ the sets that you would take the limit of to find $\limsup$. That is the sequence of all of the sups of all of the subsequences? – John O'Neil Mar 21 '17 at 4:04