1
$\begingroup$

Let $\mathcal{H^s}$ be the s-dimensional outer Hausdorff measure.

For all $t \in (0,n)$ there is none countable cover $\lbrace F_i \rbrace^\infty_{i=1}$

of $\mathbb{R^n}=\bigcup_{i=1}^{\infty} F_{i}$ with $\mathcal{H^s}(F_i)<\infty$

I used the contradiction::

If it would be possible then

$n=$dim$_H(\mathbb{R^n})=$sup dim$_H(F_i)\leq s<n$

So there can't be such a cover.

Can it be proved like this or is there something to improve?

$\endgroup$
-1
$\begingroup$

That looks fine, provided that you can prove $$\dim_H \left( \bigcup_i F_i \right) = \sup_i \dim_H(F_i).$$ It shouldn't be a difficult proof.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.