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I reading some bits and pieces of analys (some, Sakarchi and Stein, Tao). In general, if $f(0) = +\infty$ and $\forall (x \not = 0) |f(x)|<10$, would $f$ be called bounded or unbounded?

Maybe more generally, is any function in the extended real numbers considered unbounded? Or do we usually say they are all bounded, because they are bounded by $+ \infty$?

Ovi
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    -infinity and +infinity always serve als lower and upperbound, so every function to the extended reals is bounded. –  Apr 26 '18 at 07:35
  • @Math_QED So when I read a theorem which explicitly mentions bounded or unbounded, most likely the theorem is not about the extended real numbers? – Ovi Apr 26 '18 at 07:37
  • Yes,I think so. It's hard to tell without context though. –  Apr 26 '18 at 07:37
  • Ultimately, it depends on the definition you use. In metric space analysis, a function is bounded if we can enclose the image with a certain ball. We can't find a ball with finite radius that covers the extended reals, so from this definition it would seem that the answer to your question is no. –  Apr 26 '18 at 07:43
  • @Math_QED Okay. As I said I am reading bits and pieces from everywhere, so I was wondering if there is a general consensus. – Ovi Apr 26 '18 at 07:45
  • I would go with the metric space definition, if I had to choose. –  Apr 26 '18 at 07:46

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I can think of two definitions of boundedness that could be used here.

Boundedness on ordered sets, where a set is bounded if there exist an element greater than or equal to every element in the set, and same for lesser than.

This concept is useless on the extended reals, as all sets will be bounded.

Boundedness on metric spaces, where a set is bounded if it can be contained in a ball.

The extended reals is not a metric space for the usual metric on the real numbers, but there exists metrics that turn it into a metric space. But then the distance from $-\infty$ to $\infty$ must be some finite number, and the whole extended real line must be contained in a ball of this radius (any sane metric on the extended reals should have these two numbers furthest away from each other).

Again, the concept of boundedness is uninteresting, as all sets are bounded.

In general if someone talked about sets being bounded, I would therefore assume they are restricting themselves to the definition of boundedness on the reals.

For your example though, if we restrict to the reals the function will not be defined at $0$, so it would be bounded on its domain. So for us to say it was unbounded we would have to specifically define what we mean with unboundedness on the extended reals in such a way that anything containing $\infty$ would be unbounded.

Greebo
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