3
$\begingroup$

Possible Duplicate:
Continuous functions on a compact set

Let $K$ be a nonempty subset of $\mathbb{R}^n$, where $n > 1$. Why does it follows that if every continuous real-valued function defined on $K$ is bounded, then $K$ is compact?

  • 0
    I'm not sure about this, but I think this is related to the extreme value theorem, which states that every continuous real-valued function defined on a compact subset of $\mathbb{R}^n$ is bounded. Therefore, if there are no unbounded function, it'd be tempting to guess it's compact. There are some gaps to fill in here though.2011-10-14

1 Answers 1

6

If $K$ is not compact, then either it is not closed or it is not bounded. In either case, it is easy to construct a continuous unbounded real-valued function on $K$. If $K$ is not closed, take the reciprocal of the distance to a limit point. If $K$ is not bounded, take the modulus.

  • 0
    "If$K$is not closed, take the reciprocal of the distance to a limit point." - perhaps you can explicitly say that the limit point is *not in $K$*. Of course, it's clear that that's what you meant, since if we take a limit point in $K$, then the function is not even defined at that point, and hence cannot be continuous in $K$.2011-10-14