# Locally bounded argument

1. Jun 14, 2010

### ocohen

Having a hard time understanding this example from a book:

The function f(x) = 1/x is locally bounded at each point x in the set E = (0,1).
Let x \in (0,1). Take \delta_x = x/2, M_x = 2/x. Then
f(t) = 1/t <= 2/x = M_x
if
x/2 = x-\delta_x < t < x + \delta_x

This argument is false since the point 0 is an accumulation point that does not belong to (0,1). As such there is no assumption that 0 is bounded. This can be avoided by making E closed.

I don't understand this last part. Why does it matter that 0 is not bounded? I thought the whole point of locally bounded is that we can define the bound in terms of x. And we can always find a small enough interval (namely x/2) that will not include 0.

Can anyone shed some light on this?

Thanks

2. Jun 14, 2010

### Office_Shredder

Staff Emeritus
It sounds like this is supposed to be a counterexample to the statement locally bounded implies bounded. On the other hand if the domain is compact and the function is locally bounded and continuous, then the function is bounded

EDIT: In retrospect that's not very enlightening (since the locally bounded part is completely unnecessary). I'm not sure what the point is