Suppose $S$ is an abstract metric space.
Suppose $f(x):\Bbb{S} \to \Bbb{R}$. Then how what would the question be asking, if it asks "Show that $f$ is not bounded on $\Bbb{S}$"?
Is it asking me to show that $f$ isn't bounded on the codomain $\Bbb{R}$, for all terms in the domain? So I just need to show that there doesn't exist a Least upper bound/greatest lower bound, for $f(x)$.
Or does it mean for me to show that $f$ isn't bounded on the domain? Would I need to show that there is no Least upper bound or Greatest lower bound for the elements in the domain? How would I show something like greatest or least, in abstract metric spaces?
$\endgroup$2 Answers
$\begingroup$Good question. The terminology "bounded on the domain" is a little confusing at first, for the reasons you mention. It means the first thing you mention. But of course the codomain may be unbounded, although $f$ is bounded.
The reason for the terminology is that sometimes $f$ is defined on an even larger set...for instance we might have $f:\mathbb{R} \to \mathbb{R}$ and we might say "$f$ is bounded on $[0,1]$." That would mean that if we just take $x$ values in $[0,1]$, there is some value in the codomain that $f$ stays below.
$\endgroup$ $\begingroup$Function $f:X\rightarrow (Y,d)$ (where $X$ is an arbitrary set) on set $X$ is called bounded if for some $a\in X$ there exists a real number $M$ such that its distance function $d$ ("distance") is less than $M$, i.e. $$d(f(x),a)<M$$ for all $x\in X$.
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