Having a hard time understanding this example from a book:(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Locally bounded argument

Loading...

Similar Threads - Locally bounded argument | Date |
---|---|

I Upper and lower bounds of integral | Sep 8, 2016 |

Local max and min (including a domain) | Dec 4, 2015 |

Local uniform continuity of a^q | Aug 19, 2014 |

Local maxima and minima | Jun 13, 2014 |

Local min no other zeros of gradient | Apr 23, 2014 |

**Physics Forums - The Fusion of Science and Community**