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In my notes, I have written that

\begin{align} \limsup_{x\to a} f(x) & = \inf_{\delta\to0}\sup\left[f(x): \Vert x-a \Vert < \delta \right] \\ & = \inf_{n\to \infty}\sup\left[ f(x): \Vert x-a \Vert <\frac{1}{n} \right] \\ & = \lim_{n\to \infty}\sup\left[ f(x): \Vert x-a\Vert < \frac{1}{n} \right] \end{align}

Can someone help me understand the first line? I thought that the supremum is always a number, so why we are taking the infimum of a number?

Also I tried drawing a picture to illsute what the heck is going on. But basically my gist is that given an interval around $a$, as $\delta \to 0$, that supremum over that interval is the limit supremum

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the supremum is always a number, so why we are taking the infimum of a number? The supremum depends on $\delta$, and you are considering the infimum of these. – Did Dec 8 '12 at 20:59
@did, so if my $\delta = 1$, then my supremum is a number still. I don't see it – Hawk Dec 8 '12 at 21:33
See my answer. $ $ – Did Dec 8 '12 at 22:45
up vote 2 down vote accepted

For some $a$ and for every $\delta>0$, consider the set $A(\delta,a)=\left\{f(x): 0<\Vert x-a \Vert < \delta \right\}$ and the number $g(\delta,a)=\sup A(\delta,a)$.

Then, $\limsup\limits_{x\to a} f(x)$ is, equivalently, $\inf\limits_{\delta>0}g(\delta,a)$ or $\lim\limits_{\delta\to0}g(\delta,a)$. (But note that $\inf\limits_{\delta\to0}$ does not exist, only $\inf\limits_{\delta>0}$ does.)

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OKay so I understand $\limsup_{x \to a} f(x) = \lim_{\delta \to 0} g(\delta,a)$, but I don't see how $\inf_{\delta >0} g(\delta, a) = \lim_{\delta \to 0} g(\delta, a)$ – Hawk Dec 8 '12 at 22:53
This is because the function $\delta\mapsto g(\delta,a)$ is nondecreasing. – Did Dec 8 '12 at 23:25
In words, can you explain the meaning of $\inf_{\delta > 0}$ or even what it means as $\inf \{ f(x) : \Vert x - a \Vert < \delta \}$ – Hawk Dec 9 '12 at 5:01
Well, $B=\{f(x)\mid \|x-a\|\lt\delta\}$ is a set of real numbers and one considers $\inf B$. Do you know the definition of the infimum of a subset of the real line? – Did Dec 9 '12 at 8:18

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