- #1
ocohen
- 24
- 0
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
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