- #1
Useful nucleus
- 370
- 58
I'm seeking a neater proof for the following:
Let E [itex]\subseteq[/itex] R. Let every continuous real-valued function on E be bounded. Show that E is compact.
I tried to argue based on Heine-Borel theorem as follows:
E cannot be unbounded because if it is the case, define f(x)=x on E and f(x) is continuous but unbounded.
E cannot be not closed, because E=(0,1] is not closed and we can define f(x)=1/x which is continuous but still unbounded.
Thus, E has to be both bounded and closed and hence compact.
I think this argument is not very solid and I'd appreciate any hints.
Let E [itex]\subseteq[/itex] R. Let every continuous real-valued function on E be bounded. Show that E is compact.
I tried to argue based on Heine-Borel theorem as follows:
E cannot be unbounded because if it is the case, define f(x)=x on E and f(x) is continuous but unbounded.
E cannot be not closed, because E=(0,1] is not closed and we can define f(x)=1/x which is continuous but still unbounded.
Thus, E has to be both bounded and closed and hence compact.
I think this argument is not very solid and I'd appreciate any hints.