- #1

- 25

- 0

- Thread starter OB1
- Start date

- #1

- 25

- 0

- #2

- 1

- 0

in [itex]\mathbb{R}^n[/itex], you will not find any set that has a dimension [itex]>n[/itex] !! To construct a set with Hausdorf dimension [itex]0 < d < n[/itex], it suffices to construct a set [itex]E[/itex] that has a [itex]d[/itex]-Hausdorf measure [itex]0 < \mathcal{H}^{(d)}(E) < \infty[/itex]. If you know the example of Cantor-like sets in [itex]\mathbb{R}[/itex], you can adapt the idea in [itex]\mathbb{R}^n[/itex].

Last edited:

- #3

- 25

- 0

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 7K

- Last Post

- Replies
- 4

- Views
- 605

- Replies
- 7

- Views
- 6K

- Replies
- 2

- Views
- 4K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 3K