# Proof that $\liminf_{x\to\infty}f(x) \leq \limsup_{x\to\infty}f(x)$.

I can't understand this proof from my old lecture notes. $\liminf$ is defined as: \begin{align*} \liminf_{z\to\infty}f(z) = \inf_{x< y} \sup_{y< z}f(z) \end{align*} and $\limsup$ is defined similarly. We have already shown that if $f$ is an increasing (decreasing) function then $\sup f$ ($\inf f$) is unique, if it exists, hence the definition of liminf is safe. We have also shown that \begin{align*} \inf_{y<z} f(z) \leq \sup_{y<z} f(z). \end{align*} We assume that both $\liminf f$ and $\limsup f$ exist. The proof is presented as follows: \begin{align*} \inf_{y<z} f(z) &\leq \sup_{y<z} f(z)\\ \sup_{x<y} \inf_{y<z} f(z) &\leq \sup_{x<y} \sup_{y<z} f(z)\\ \inf_{x<y} \inf_{y<z} f(z) &\leq \inf_{x<y} \sup_{y<z} f(z) \end{align*} I don't see how this proves the theorem. Could somebody please explain the conclusion?

P.S. At this stage we do not have $\lim$ in our toolbox.

• Are you sure your definition is correct? Because it makes very little sense, with three variables where $x$ is only ever under the expression... – 5xum Apr 18 '16 at 10:16
• Yes, in this context it makes because $\sup_{y< z}f(z)$ is decreasing in $y$ and hence $\inf_{x< y} \sup_{y< z}f(z)$ is independent of $x$, if it exists. The theorem assumes that $\liminf_{x\to\infty} f(x)$ exists so we are safe. – lampishthing Apr 18 '16 at 10:21
• You are missing my point. The point is that $\sup_{y<z}f(z)$ is independent of $x$, therefore $\inf_{x<y}\sup_{y<z} f(z)$ is the infimum of a constant, which means $\inf_{x<y}\sup_{y<z} f(z)=\sup_{y<z} f(z)$, which is a function of $y$. – 5xum Apr 18 '16 at 10:40

The correct definitions should be $$\liminf_{z\to\infty} f(z)=\sup_y\inf_{z\ge y}f(z),\qquad\limsup_{z\to\infty} f(z)=\inf_{y}\sup_{z\ge y}f(z).$$ Here is two-line proof: $\;\inf_{z\ge y} f(z) \leq \sup_{z\ge y} f(z)$, hence \begin{align*} \inf_{z\ge y} f(z) &\leq \inf_y\sup_{z\ge y} f(z)&&\qquad \text{by definition of the g.l.b.}\\ \sup_y\inf_{z\ge y} f(z) &\leq \inf_y\sup_{z\ge y} f(z)&&\qquad \text{by definition of the l.u.b.} \end{align*}