$ A \subset \Bbb R$ is bounded and not compact, exhibit $f: A \to \Bbb R$ that isn't bounded

Question

If $ A \subset \Bbb R$ is bounded (but arbitrary -- it doesn't have to be an interval or even connected) and not compact, can somebody please provide an example of a function $f: A \to \Bbb R$ that isn't bounded? I'm having trouble seeing how it's possible...

Answer 1

If $A$ is just infinite and you have no regularity assumptions then it is trivial: enumerate any countable subset of $A$, say $x_n$, and set $f(x_n)=n$. Set $f$ to be whatever you want on the rest of $A$. This would work even if $A$ were compact. I only point this out to emphasize to you that it is important to include all your hypotheses when you pose a problem.

The more interesting question is to make $f$ continuous. For that, pick an element of the "strict boundary" of $A$ i.e. $\overline{A} \setminus A$, say $x_0$, and define $f(x)=|x-x_0|^{-1}$.