An example of a function that attains the value infinity on R?

[AxiomOfChoice]

An example of a function that attains the value "infinity" on R?

I'm reading a couple of books on introductory measure theory (Royden, Stein-Shakarchi), and both of them talk about functions that can possibly attain the value $\infty$. But they don't define exactly what this means, or give examples. So can someone list some examples? Does $1/x$ attain the value of $\infty$ at $x=0$? Does $1/x^2$? In this sense, is the latter function continuous at $x=0$, whereas the former isn't?

 

[Hurkyl]

But they don't define exactly what this means

It means, there exists some real number r such that $f(r) = +\infty$.

Your text (surely) has explicitly defined the extended real numbers by now. They are a number system that contains every real number, and two additional numbers they call $+\infty$ and $-\infty$. There is nothing syntactically special about these two infinite numbers -- they are extended real numbers just like any other extended real numbers.

Does $1/x$ attain the value of $\infty$ at $x=0$?

No.

Does $1/x^2$?

Yes and no.

Taking this expression literally as expressing a partial function in the extended real variable x, it is undefined at x=0.

However, in continuous mathematics, it is common to take the "continuous extension" of a function. 1/x² has a limit at x=0, and we can define a new (continuous!) function to be 1/x² everywhere else, and to be $+\infty$ at x=0.

It is common to do this continuous extension without explicitly stating it. And in a context like this, I would assume that is what was meant unless it was explicitly stated otherwise. I would expect a good introductory text to at least mention that we often do this...

Anyways, you will eventually be tweaking the definition of "function" so that its values at individual points don't really matter -- so it doesn't really matter if 1/x² is defined at x=0 and what value it might have.


If we desired, we could extend 1/x by defining it to be $+\infty$ at x=0, but it still wouldn't be continuous there.


Note that continuous extension happens in a different way too -- e.g. it would compel us to say $\mathrm{arctan}(+\infty) = \pi/2$ and $\mathrm{arctan}(-\infty) = -\pi/2$.


P.S. Just FYI, I am taking care to write $+\infty$ for the positive infinite number. My reason, which is probably irrelevant to you, is to distinguish it from the projective real numbers which just has one infinite number, which I call $\infty$.

 

[HallsofIvy]

The reason why 1/x does NOT take on the value $\infty$ at x=0, even in the extended real numbers, while 1/x² does, is that there exist (negative) numbers arbitrarily close to 0 for which 1/x is an arbitarily large negative number, while there are also numbers arbitrarily close to 0 for which 1/x is an arbitrarily large positive number. For x close to 0, 1/x² is always positive.

 

[AxiomOfChoice]

Thanks for your help.

Can anyone else provide examples of functions that attain the value [tex]\infty[/itex] at some point in the non-extended real numbers?

 

[g_edgar]

Example...

$$f(x) =\begin{cases} 0 \text{ if }x<1\\ \infty \text{ if } 1 \le x \le 2\\ 4 \text{ if } x > 2\end{cases}$$