User:IssaRice/Taking inf and sup separately

This page describes a trick that is sometimes helpful in analysis.

Satement

Let $A$ and $B$ be bounded subsets of the real line. Suppose that for every $a\in A$ and $b\in B$ we have $a\geq b$. Then $\inf(A)\geq \sup(B)$.

Actually, do $A$ and $B$ have to be bounded? I think they can even be empty!

Proof

Let $a\in A$ and $b\in B$ be arbitrary. We have by hypothesis $a\geq b$. Since $b$ is arbitrary, we have that $a$ is an upper bound of the set $B$, so taking the superemum over $b$ we have $a\geq \sup(B)$ (remember, $\sup(B)$ is the least upper bound, whereas $a$ is just another upper bound). Since $a$ was arbitrary, we see that $\sup(B)$ is a lower bound of the set $A$. Taking the infimum over $a$, we have $\inf(A)\geq \sup(B)$, as required.

Applications

liminf vs limsup

(Notation from Tao's Analysis I.)

Let $(a_{n})_{n=m}^{\infty }$ be a sequence of real numbers. Let $L^{-}:=\liminf _{n\to \infty }a_{n}$ and let $L^{+}:=\limsup _{n\to \infty }a_{n}$. Then we have $L^{-}\leq L^{+}$.

Consider the sequences $(a_{N}^{-})_{N=m}^{\infty }$ and $(a_{N}^{+})_{N=m}^{\infty }$ defined by $a_{N}^{-}:=\inf(a_{n})_{n=N}^{\infty }$ and $a_{N}^{+}:=\sup(a_{n})_{n=N}^{\infty }$.

Now consider the sets $A:=\{a_{N}^{+}:N\geq m\}$ and $B:=\{a_{N}^{-}:N\geq m\}$. If we can show that $a_{j}^{+}\geq a_{k}^{-}$ for arbitrary $j,k\geq m$, then we can apply the trick to these sets to conclude that $L^{+}=\inf(a_{N}^{+})_{N=m}=\inf(A)\geq \sup(B)=\sup(a_{N}^{-})_{N=m}=L^{-}$.