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Wikipedia has the definition for metric space, but it doesn't have a definition for extended real-field.

It just says, 'Limit inferior and limit superior of a real function are always well-defined in $\overline{\mathbb{R}}$'.

In that sense, I don't think the definition for metric space covers real-valued functions. (Isolated points)

What's the definition it would be?

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Related (to some extent) About the notion of limsup and liminf – Martin Sleziak Dec 6 '12 at 7:32
up vote 0 down vote accepted

You wrote:

In that sense, I don't think the definition for metric space covers real-valued functions. (Isolated points)

If you are asking what metric you should take on $\mathbb R\cup\{\pm\infty\}$ so that the definition for metric spaces yields the correct definition for real function, then the answer is: Identify $\mathbb R\cup\{\pm\infty\}$ with $[0,1]$ via some order-preserving bijection. Transfer the metric from $[0,1]$ to $\mathbb R\cup\{\pm\infty\}$ using this bijection. Then the definition given in Wikipedia article for function to an ordered metric space gives as a special case the definition for functions to $\overline{\mathbb R}$.

Here is link to the current revision of the Wikipedia article linked above. (Just in case it changes substantially in the future.)

EDIT: (In response to the comment asking about $\limsup\limits_{x\to\infty} f(x)$ for a function from $\mathbb R$ to $\mathbb R$.)

$\limsup_{x\to a} f(x)$ is defined even if the value of $f$ at the point $a$ is not defined. If we follow the definition from Wikipedia, we should look at punctured neighborhood of $\infty$. The balls around $\infty$ are the intervals of the form $(x,\infty)$. So we get
$$\limsup\limits_{x\to\infty} f(x) = \lim\limits_{x\to\infty} \sup\{f(a); a>x\} = \inf\limits_{x\in\mathbb R} \sup\{f(a); a>x\}.$$

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Then, what's the meaning of "$\limsup_{x\to\infty} f(x)$" where $f$ is a map from $\mathbb{R}$ to $\mathbb{R}$? – Katlus Dec 6 '12 at 20:08
Also, wikipedia says, "$Y$ should be an ordered set". Does this mean $Y$ should be well-ordered? Or just fully-ordered? I guess, in ZF, to define limit inferior of a function from $X$ to $Y$ where $X,Y$ are topological spaces, $Y$ should be Dedekind-complete field. Am i correct? – Katlus Dec 6 '12 at 20:23
You need that the suprema appearing in the definition of $\limsup$ exist. A complete lattice should be enough; perhaps this can be weakened. (I am not sure.) Least-upper-bound property (Dedekind completeness) should work. To be honest, I never needed $\limsup$ in something different than $\mathbb R$ and $\mathbb R\cup\{\pm\infty\}$. – Martin Sleziak Dec 6 '12 at 20:31

Say one considers a function $f:(0,+\infty)\to\mathbb R$ and one wants to determine the limsup/liminf of $f$ at $0$. Then one would define $g(t)=\sup\{f(x)\mid0\lt x\lt t\}$, $h(t)=\inf\{f(x)\mid0\lt x \lt t\}$, and $$ \limsup\limits_{x\to0}f(x)=\inf\{g(t)\mid t\gt0\}=\lim\limits_{t\to0}g(t), $$ and $$ \liminf\limits_{x\to0}f(x)=\sup\{h(t)\mid t\gt0\}=\lim\limits_{t\to0}h(t). $$ Example: $f(x)=\sin(1/x)$ for every $x\gt0$, then $g(t)=+1$ and $h(x)=-1$ for every $t\gt0$ hence $\limsup\limits_{x\to0}f(x)=+1$ and $\liminf\limits_{x\to0}f(x)=-1$.

Similar but more involved: if $f(x)=(2+x^2\cos(1/x))\sin(1/x)$ for every $x\gt0$, then $\limsup\limits_{x\to0}f(x)=+2$ and $\liminf\limits_{x\to0}f(x)=-2$.

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Just consider the function

$$ f(x) = \frac{|x|}{x}.$$

Can you determine the limsup and the liminf as $x \to 0$ of this function? See here for the definition.

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